Let's talk with probability

1. With the reference to the pair of dice one red and one green throwing experiment, calculate the probability:

    a) The sum of the numbers is divisible by 4:

        We need to first list out all the possible combinations that yield a sum divisible by 4. The minimum sum we can get from two dice is 2 and the maximum sum is 12. Between 2 and 12 that are divisible by 4 are 4 and 8. 

        

• SUM = 4 = (1, 3), (2, 2), (3, 1)
• SUM = 8 = (2, 6), (3, 5), (4, 4), (5, 3), (6, 2)

    Counting up all the combinations, we have 3 + 5 = 8. So, the probability that the sum of the numbers on the two dice is divisible by 4 is = 8 / 16 = 2 / 9.

    Required probability = 2 / 9.

    b) The number on the green die is greater than the number on the red die:

    Let's know all outcomes where the number on the green die is greater than the number on the red die.

• If red shows 1, then green shows 2, 3, 4, 5, 6. So, it has 5.
• If red shows 2, then green shows 3, 4, 5, 6. So, it has 4.
• If red shows 3, then green shows 4, 5, 6. So, it has 3.
• If red shows 4, then green shows 5, 6. So. it has 2.
• If red shows 5, then green shows 6. So, it has 1.
• If red shows 6, then green shows 0. So, it has 0.

    Now, adding all the possibilities: 5 + 4 + 3 + 2 + 1 = 15

The probability P (G > R) = 15 / 36 = 5 / 12.

Required probability = 5 / 12.

    c) The sum of the numbers is divisible by either 4 or 3 or both 4 and 3:

• Possible sums from two dice range from 2 to 12. 
• Sums divisible by 4 from the range 2 to 12 are 4, 8, 12.
• Sums divisible by 3 from the range 2 to 12 are 3, 6, 9, 12.
• For 4, we have 7 possibilities and for 3, we have 12 possibilities.

So, we need to subtract the ways we can get 12. Therefore, the total number is 7 + 12 - 1 = 18.

    d) The sum of the numbers is divisible by 3 and 4 both:

    To determine the total number of possible outcomes when rolling two dice, each die has 6 sides. There are 2 dice, the total number of possible outcomes is 6 * 6 = 36. There is only 1 outcome where the sum is 12 it divisible by 3 and 4.

    Required probability = 1 / 36.

    e) The sum of the numbers is odd, and the first die shows an even number:

    We need to find both events A as the sum of the numbers is odd, and the event B as shows the even number. 

• The even numbers are 2, 4, 6.
• The probability that the red die shows an even number is.

    P(B) = 3 / 6 = 1 / 2.

    

    If the red die shows 2, the green die needs to show a 1, 3, 5. If the red shows 4, then the green needs to show 1, 3, 5. If red shows 6, then also same. The probability that the sum is odd is consistently as 1 / 2.

    P (A and B) = 1 / 2 * 1 / 2 = 1 / 4.

    Required probability = 1 / 4.

    f) The sum of the numbers is less than five or the same number occurs on both the dice:

    The combinations we have, 1+2, 2+1, 1+3, 3+1, 1+1, 2+2, 3+3, 4+4, 5+5, 6+6. The total as 10 outcomes. There are 6 * 6 = 36 possible outcomes when rolling two dice. the probability of the event is 

P (Sum < 5 or same number on both dice) = 10 / 36 = 5 / 18.

    Required probability = 5 / 18.

    g) The sum of the numbers is 3 or 11:

    The probability of a Sum of 3 as 2 favorable outcomes and the probability of a Sum of 11 as 2 favorable outcomes. The total favorable outcomes as 4. 

    Required probability = 4 / 36 = 1 / 9.

2. There are 30 balls in an urn. The balls are numbered from 1 to 30, one number per ball. If one of the balls is drawn at random

    a) The probability that the number on it is divisible by 2 and 3:

    Let's find the numbers between 1 and 30 that are divisible by 6 as 6, 12, 18, 24, 30. They are 5 numbers. The probability that the number on the drawn ball is divisible by both 2 and 3 is 1 / 6.

    b) The probability that the number on it is a perfect square:

    

    They are 1 Square, 2 Square, 3 Square, 4 Square, 5 Square. There are 5 balls with numbers that are perfect squares. The probability that the number on a randomly drawn ball from the urn is a perfect square is 5 / 30 = 1 / 6.

    c) The probability that the number on it is a 2-digit number:

    The smallest two-digit number is 10 and the largest two-digit number is 30. There are 30 - 10 + 1 = 21.  The total balls as 30.

    Required probability = 21 / 30 = 7 / 10.

    d) The probability that the number on it is a prime number:

    Let's identify the prime numbers between 1 and 30 as 2, 3, 5, 7, 11, 13, 17, 19, 23, 29. They are 10 prime numbers between 1 and 30. 

    Required probability = 10 / 30 = 1 / 3.

    e) The probability that the number on it is a prime number that is 2 more than another prime number:

    To find the probability that the number on the drawn ball is prime number that is 2 more than another prime number, we first need to identify the prime numbers between 1 and 30 that satisfy this criterion. These are often called "Twin primes". They are (3, 5), (11, 13), (17, 19), (29, 31). We are considering 29 only hear. So, there are 4 favorable outcomes.

    Required probability = 4 / 30 = 2 / 15.

3. A letter of the English alphabet is chosen at random. 

    a) The probability that the letter so chosen is a consonant:

    There are 26 letters in the English alphabet. It has 21 Consonants are B, C, D, F, G, H, J, K, L, M, N, P, Q, R, S, T, V, W, X, Y, Z. 

    Required probability = 21/26.

    b) The probability that the letter so chosen precedes r (in order):

    To determine the number of letters that precedes "r" in order. we count as a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q. There are 17 letters.

    Required probability = 17/26.

    c) The probability that the letter so chosen precedes x and is a vowel:

    The vowels are A, E, I, O, U. The letters that precede 'x', the vowels are A, E, I, O, U.

    Required probability = 5/26.

    d) The probability that the letter so chosen follows w and is a vowel:

    The letter that follows "w" in the English alphabet is "x". The vowels in the English alphabet are a, e, i, o, u. 

    Required probability = 0.

    e) The probability that the letter so chosen follows p and precedes y and is a vowel:

    The only letter that is between 'p' and 'y' is q, r, s, t, u, v, w, x. The only vowel is 'u'.

    Required probability = 1 / 26.

4. In a single throw of 3 dice, find the probability of getting the same number on all the three dice.

    The favorable outcomes for getting the same number on all three dice are (1, 1, 1), (2, 2, 2), (3, 3, 3), (4, 4, 4), (5, 5, 5), (6, 6, 6). There are 6 favorable outcomes in total.

    Required probability = 6 / 216 = 1 / 36.

5. In a single throw of three dice, find the probability of getting a sum of 5 and at least 5.

    For three dice, the combinations that result in a sum of 5 are (1, 1, 3), (1, 3, 1), (3, 1, 1), (1, 2, 2), (2, 1, 2), (2, 2, 1), (1, 1, 4), (1, 4, 1), (4, 1, 1), (2, 3, 0). Zero is not possible. There are 9 combinations to get a sum of 5. 

    For three dice, the combinations that result in a sum of at least 5 are (1, 1, 1), (1, 2, 1), (1, 1, 2), (2, 1, 1). The total as 216. 

    Required probability for sum of 5 = 1 / 24.

    Required probability for at least 5 = 53 / 54.

6. From pack of playing cards, 2 cards are drawn at random. Find the probability that the one is a king and other is a queen.

    Total number of cards are 52, The number of kings in the deck are 4 and the number of the queens in the deck are 4. The probability that the first card is a king as 4/52. Once a king has been drawn, there are now 51 cards left in the deck. The probability that the second card is a Queen as 4/51. The probability that the first card is a King, and the second card is a Queen = (1/13) * (4/51) = 4/663.

    The probability that the first card is a queen as 4/52 = 1/13. Once a queen has been drawn, there are now 51 cards left in the deck. The probability that the second card is a king as 4/51. The probability that the first card is a queen, and the second card is a king = (1/13) * (4/51) = 4/663.

    Required probability = 4/331.5

7. Two cards are drawn at random from a well-shuffled pack of 52 cards. Show that the chance of drawing two aces is (1/221).

    The number of aces in the deck are 4, the probability that the first card is an ace = 4/52 = 1/3. There are now 3 aces left in the deck and 51 total cards remaining. The probability that the second card is an ace = 3/51 = 1/17. The combined probability is the product of the two individual probabilities as 1/221.

    Required probability = 1/221.

8. A bag contains 6 red, 5 white and 4 black balls. If two balls are drawn, find the probability that none of them is red.

    The total number of balls are 6 red + 5 white + 4 black = 15 balls. There are 9 balls that are not red (5 white + 4 black) out of a total of 15. The probability that the first ball drawn is not red = 3/5. There are now 8 non - red balls left and 14 balls in total. The probability that the second ball drawn is not red = 4/7. The combined probability is the product of the two individual probabilities. 

    Required probability = 12/35.

9. A bag contains 7 red balls and 5 white balls. 4 balls are drawn at random. what is the probability that all of them are red?

    The total number of balls are 7 red + 5 white = 12 balls. The probability that the first ball drawn is red as 7/12. The probability that the second ball drawn is red as 6/11. The probability that the third ball drawn is red = 5/10 = 1/2. The probability that the fourth ball drawn is red as 4/9. The probability of drawing four balls consecutively and all of them being red from the given bag is 7/33.

    Required probability = 7/33.

10.  An urn contains 9 balls, two of which are red, three blue and four black. Three balls are drawn from the urn at random.

    a) The probability that the balls are of different colors:

        

        It means one red, one blue, one black is drawn. The total favorable outcomes for this as 2*3*4 = 24. The probability for the balls to be of different colors is 24/84 = 2/7.

    b) The probability that the two balls are of the same color and third is of different color:

        The total favorable outcomes are 3 + 4 + 6 + 12 + 12 + 18 = 55. 

        P (2 Same color, 1 different) = 55/84.

    c) The probability that the three balls are of the same color:

        The total favorable outcomes are 1 + 4 = 5.

        P (all same color) = 5/84.

11. In a lottery of 50 tickets numbered 1 to 50, two tickets are drawn simultaneously.

    a) The probability when both the tickets drawn have prime numbers:

        The prime numbers between 1 and 50 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47. There are 15 total. The number of ways to draw 2 prime tickets from 15 as C (15,2) = 15! / (2! * (15-2)!) = 15! / (2! *13!) = (15*14)/2 = 105 ways. The total number of ways to draw 2 tickets out of 50 is C (50,2) = 50! / (2! * (50 - 2)!) = 50! / (2! * 48!) = (50*49)/2 = 1225 ways.

        Required probability = 105/1225 = 3/35.

    b) The probability when none of the tickets drawn has prime number:

        There are 15 prime numbers between 1 and 50. The number of non-prime tickets as 35. The ratio of the number of non-prime tickets to the total number of tickets P (1st ticket is not prime) = 35/50 = 7/10. There are 49 tickets remaining and 34 of them are not prime as 34/49. 

        P (Neither ticket is prime) = (7/10) * (34/49)

        Required probability = 238 / 490 = 119 / 245.

12. What is the probability of getting 9 cards of the same suit in one hand at a game of bridge?

        We have 52 cards, which consists of 4 suits: diamonds, hearts, clubs, spades. Each has 13 cards. When playing bridge, each player is dealt 13 cards. We choose 1 suit out of 4 for the 9 cards. 

    = C (4,1) = 4.

    

    Then we choose 9 cards out of the 13 of those suits = C (13, 9) = 13! / 9!4!

    Again, we need to choose 4 cards from the remaining 39 cards (52-13) = C (39, 4) = 39! /4!35!

Now, we can do the total number of ways to select the desired 13 cards = C (4,1) * C (13,9) * C (39, 4).

The total number of ways to select any 13 cards from a deck of 52 cards as C (52, 13) = 52! / 13! 39!

    Required probability = C (4,1) * C (13,9) * C (39,4) / C (52, 13).

13. The number 1,2,3,4,5,6 is written in slips of paper and two of these slips are drawn.

    For the pair of numbers chosen from the set (1,2,3,4,5,6), there are C (6,2) possible combinations.

    C (6, 2) = 6! / 2!4! = 15.

    a) The probability that the sum of the numbers drawn is 9:

        The pairs (3,6) and (4,5) sum up to 9. There are 2 outcomes = 2/15.

        

    b) The probability that the sum of the numbers drawn is 5 or less:

        The pairs (1,2), (1,3), (1,4), (2,3) sum up to 5 or less. There are 4 outcomes = 4/15.

    c) The probability that one of the numbers drawn is odd or greater than 3:

        Let's check the pairs where at least one number is odd are (1,2), (1,3), (1,4), (1,5), (1,6), (2,3), (2,5), (3,4), (3,5), (3,6), (4,5), (5,6). The pairs where at least one number is greater than 3 are (4,2), (4,6), (6,2). We have to subtract the (4,5) and (1,6). So, the final outcomes are 13. 

        Required probability = 13/15.

    d) The probability that the numbers drawn are both odd:

        The pairs where odd is (1,3), (1,5), (3,5).

        Required probability = 3/15.

14. A fair coin and a fair die are tossed, and a card is randomly selected from a standard deck of 52 playing cards.

    a) The probability that the coin, die and card, respectively show a head, a5, and the king of diamonds:

        The probability of getting a head as 1/2, The probability of getting a 5 on the die as 1/6, The probability of getting the king of diamonds as 1/56. 

        Required probability = 1/2 * 1/6 * 1/52 = 1/624.

    b) The probability that the coin, die and card, respectively show a head, a 5, and king:

        The probability of getting a head as 1/2, The probability of getting a 5 on the die as 1/6. The probability of getting a king from the deck as 1/13. 

        Required probability = 1/2 * 1/6 * 1/3 = 1/156.

    c) The probability that the coin, die and card, respectively show a tail, a 3 or 5, and a king:

        The probability of getting a tail as 1/2, The probability of getting a 3 or 5 on the die as 1/3, The probability of getting a king as 1/13. 

        Required probability = 1/2 * 1/3 * 1/13 = 1/78.

    d) The probability that the coin, die and card, respectively show a tail, an odd number, and a heart:

        The probability of getting a tail as 1/2, The probability of getting an odd number on the die as 1/2, The probability of getting a heart from the deck as 1/4.

        Required probability = 1/2 * 1/2 * 1/4 = 1/16.

15. Two different digits are chosen at random from the set 1,2, 3, ......, 8. Show that the probability that the sum of the digits will be equal to 5 is same as the probability that their sum will exceed 13, each being 1/14. Also show that the chance of both digits exceeding 5 is 3/28.

    Given set of numbers as 1, 2, 3, 4, 5, 6, 7, 8.

     There are 4 combinations are (1,4), (2,3), (3,2), (4,1). There are C (8, 2) ways to choose 2 numbers from the set of 8, where C (n, r) represents combinations as C (8, 2) = 8! /2!6! = 28. The probability that the sum of two numbers is 5. The probability that the sum of two number is 5 as 1/7. 

    The probability that the sum of the two numbers will exceed 13, the combinations are (6,8), (7,7), (7,8), (8,6), (8,7). There are 5 combinations. The probability that the sum of two numbers exceeds as 13.

        P (Sum > 13) = 5/28 = 1/7 * 5/5 = 1/14 * 2 = 1/7.

    

    The numbers greater than 5 are 6, 7, and 8. The combinations where both numbers are greater than 5 are (6,7), (6,8), (7,6), (7,8), (8,6), (8,7). There are 6 combinations that exceeds.

    P (both numbers > 5) = 6/28 = 3/14.

    Hence, the chance that both digits exceed 5 is 3/14.

16. A die is loaded in such a manner that for n = 1,2,3,4,5,6 the probability of the face marked n, landing on top when the die is rolled is proportional to n. Find the probability that an odd number will appear on tossing the die.

    The probability P (n) of the face marked n (where n ranges from 1 to 6). The sum of the probabilities for all sides of the die should equal 1 P (1) + P (2) + P (3) + P (4) + P (5) + P (6) = 1. By Substitution, we get k (1) + k (2) + k (3) + k (4) + k (5) + k (6) = 1.

    k = 1/21.

    The probability that an odd number will appear is.

    P (odd) = P (1) + P (3) + P (5)

    P (odd) = 1+3+5 / 21 

    Required probability = 3/7.

17. A bag contains 50 tickets numbered 1,2,3, ......., 50 of which five are drawn at random and arranged in ascending order of the magnitude (x1 < x2 < x3 < x4 < x5). What is the probability that x3 = 30?

    Let's take x3 = 30.

    The third ticket is 30. We need to consider the number of ways we can choose 2 tickets form the first 29 numbers and 2 tickets from the last 20 numbers. The number of ways to choose 2 tickets from the first 29 as C (29,2). The number of ways to choose 2 tickets from the last 20 as C (20,2). The total combinations for this scenario as C (29, 2) * C (20, 2). The total number of ways to draw 5 tickets from 50, the combination C (50, 5). 

    Required probability = C (29, 2) * C (20, 2) / C (50, 5).

18. In a random arrangement of the letter of the word 'STATISTICS', find the probability that all the vowels come together.

    The word "STATISTICS" has 10 letters as S (3), T (3), A (1), I (2), C (1). The total number of arrangements without any restrictions as 10! / 3! 3! 1! 2! 1!

    The vowels are A and I, there are 3 vowels A I I. The number of arrangements of these 8 entities are 8! / 3! 3! 1!

    The total number of ways in which the vowels come together as 

    8! / 3! 3! 1! * 3! / 2!

    Required probability = 8! / 3! 3! 1! * 3! / 2! / 10! / 3! 3! 1! 2! 1!

19. A factory needs two raw materials, say X and Y. The probability of not having an adequate supply of material X is 0.06, whereas the probability of not having an adequate supply of material Y is 0.04. A study shows that the probability of a shortage of both X and Y is 0.02. What is the probability of the factory being short of either material X or Y?

    P (Short of X or Y) = 0.06 + 0.04 - 0.02

    P (Short of X or Y) = 0.08.

20. A carpenter has a tool chest with two compartments, each one having a lock. He has two keys for each lock, and he keeps all four keys in the same ring. His habitual procedure in opening a compartment is to select a key at random and try it. If it fails, he selects one of the remaining three and tries it and so on. Show that the probability that he succeeds on the first, second and third try is 1/2, 1/3, 1/6, respectively.

    The probability that the key he selects is correct on the first try is 1/2. The probability that he fails on the first try but succeeds on the second try can be found by considering the sequences of choice as 2/3. The probability that he fails the first time but succeeds the second time as 1/3. The probability that he fails the first two times but succeeds on the third try is 1/6.

Learn some topics about Probability!

   According to Ya-Lin Chou "Probability is the science of decision-making with calculated risks in the face of uncertainty.

    If a random experiment or a trail results in 'n' exhaustive, mutually exclusive and equally likely outcomes, out of which m are favorable to the occurrence of an event E, then the probability 'p' of occurrence of E, usually denoted by P(E), is given by: 

                

              

    

    There are number of events in day-to-day life about which one is not sure whether it will occur or not. But there is a curiosity to know what chance their event is to occur. For example, one may be interested to estimate profit or loss. The numerical evaluation of chance factor of an event is known as probability. An event is the collection of possible outcomes which may be favorable to an happening out of total outcomes which may be count or list out. The result of a random experiment is said to be outcome.  If in each trail of an experiment conducted under identical conditions, the outcome is not unique, but may be one possible outcome, then it is said to be random experiment. An event having only one sample point is called simple event. Two events A and B are said to be equal if A subset B and B subset A. If A, B and C are 3 events as A subset B and B subset C, it implies that A subset C. It is said to be transitivity of events. An event which is not simple is said to be compound event. If two or more events joined by the conjunction 'or' are called derived events. An event which is certain not to occur is called an impossible event. Events are said to be mutually exclusive or incompatible if the happening of any of them precludes the happening of all the others, if no two or more of them can happen simultaneously in the same trail. Outcomes of trail are said to be equally likely if taking into consideration all the relevant evidence, there is no reason to expect one in preference to the others. Several events are said to be independent if the happening of an event is not affected by the supplementary knowledge concerning the occurrence af any number of the remaining events.

1. Some important theorems on probability:

1.1 Addition theorem of probability

Statement: If A and B are any two arbitrary events in the sample space and are not disjoint, then P (A Union B) = P (A) + P(B) - P (A Intersection B) 

Proof: From the Venn diagram, we have 

1.2 Multiplication theorem of probability

Statement: For two events A and B, 

                                

    Where P (B | A) represents conditional probability of occurrence of B when the event A has already happened and P (A | B) is the conditional probability of happening of A, given that B has already happened.

Proof: In the usual notations, we have

   

 For the conditional event A | B, the favorable outcomes must be one of the sample points of B, i.e., for the event A | B, the sample space is B and out of the n(B) sample points, n(A intersection B) pertain to be occurrence of the event A. Hence

1.3 Baye's Theorem.

Statement: If E1, E2, E3, .......En are mutually disjoint events with P(Ei) is not equals to 0, (i = 1, 2, ......, n), then for any arbitrary event E which is a subset of

Ei such that P(E) > 0, 

Proof: 

Some problems:

1. Two unbiased dice are thrown.

    In a random throw of two dice, since each of the six faces of one die can be associated with each of six faces of the other die, the total number of cases is 6 * 6 = 36, as given below:

       (1, 1), (2, 1), (3, 1), (4, 1), (5, 1), (6, 1), (1, 2), (2, 2), (3, 2), (4, 2), (5, 2), (6, 2), (1, 3), (2, 3), (3, 3), (4, 3), (5, 3), (6, 3), (1, 4), (2, 4), (3, 4), (4, 4), (5, 4), (6, 4), (1, 5), (2, 5), (3, 5), (4, 5), (5, 5), (6, 5), (1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 6).

    Here, the expression, say (i, j) means that the first die shows the number i and the second die shows the number j. Obviously, (i, j) is not equals to (j, i) if i is not equals to j.

    Therefore, Exhaustive number of cases (n) = 36.

   

    a) The favorable cases that both the dice show the same number are:

  

                (1, 1), (2, 3), (3, 3), (4, 4), (5, 5) and (6, 6), i.e., m = 6.

        Therefore, Probability that has two dice show the same number = 6 / 36 = 1 / 6.

    b) The favorable cases that the first die shows 6 are:

               (6, 1), (6, 2), (6, 3), (6, 4), (6, 5) and (6, 6), i.e., 6 in all.

        Therefore, Probability that the first die shows 6 = 6 / 36 = 1 / 6.

    c) The cases favorable to getting a total of 8 on the two dice are:

               (2, 6), (3, 5), (4, 4), (5, 3), (6, 2), i.e., m = 5.

        Therefore, Probability that total of numbers on two dice is 8 = 5 / 36.

    d) The cases favorable to getting a total of more than 8 are:

               (3, 6), (6, 3), (4, 5), (5, 4), (4, 6), (6, 4), (5, 5), (5, 6), (6, 5), (6, 6), i.e., m = 10.

        Therefore, Probability that the total of numbers on two dice is greater than 8 = 10 / 36 = 5 / 18.

    e) The total of the numbers on the dice is 13: This is an example of an impossible event, since the maximum total can be 6 + 6 = 12. Therefore, the required probability is 0.

2. What is the chance that a leap year selected at random will contain 53 Sundays?

    In a leap year, there are 52 complete weeks and 2 days over. The following are the possible combinations for these two 'over' days.

•  Sunday and Monday
• Monday and Tuesday
• Tuesday and Wednesday
• Wednesday and Thursday
• Thursday and Friday
• Friday and Saturday
• Saturday and Sunday.

    In order that a leap year selected at random should contain 53 Sundays, one of the two over days must be Sunday. Since out of the above 7 possibilities, 2, to this event.

    Required probability = 2 / 7.

3. n persons are seated on n chairs at a round table. Find the probability that two specified persons are sitting next to each other.

    n persons can be seated in n chairs at a round table in (n - 1)! ways, the exhaustive number of cases = (n -1)!

    

    Assuming the two specified persons A and B who sit together as one, we get (n -1) persons in all, who can be seated at a round table in (n - 2)! ways. Further, since A and B can interchange their positions in 2! ways, total number of favorable cases of getting A and B together is (n-2)! * 2!

    Required probability = (n - 2)! * 2! / (n - 1)! = 2 / n - 1. 

4. A and B throw with three dice; if A throws 14, find B's chance of throwing a higher number.

    To throw higher number than A, B must throw either 15 or 16 or 17 or 18. Now a throw amounting to 18 must be made up of (6, 6, 6), which can occur in one way; 17 can be made up of (6, 6, 5), which can occur in 3! / 2!1! = 3 ways; 16 may be made up of (6, 6, 4), and (6, 5, 5), each of which arrangements can occur in 3! / 2!1! = 3 ways; 15 can be made up of (6, 4, 5), or (6, 3, 6), or (5, 5, 5), which can occur in 3! 3 and 1 way, respectively.

    The number of favorable cases = 1 + 3 + 3 + 3 + 6 + 3 + 1 = 20.

In a random throw of 3 dice, the exhaustive number of cases = 6 * 6 * 6 = 216.

    Required probability = 20 / 216 = 5 / 54.

5. An urn contains 4 tickets numbered 1, 2, 3, 4 and another contains 6 tickets numbered 2, 4, 6, 7, 8, 9. If one of the two urns is chosen at random and a ticket is drawn at random from the chosen urn, find the probabilities that the ticket drawn bears the number (i) 2 or 4, (ii) 3, (iv) 1 or 9.

    (i) p = P(I) + P(II) = 1/2 * 2/4 + 1/2 * 2/6 = 5/12

    (ii) 1/2 * 1/4 + 1/2 * 0 = 1/8

    (iii) 1/2 *1/4 + 1/2 * 1/6 = 5/24.

Probability

 

1. The outcome of tossing a coin is simple event.
2. Classical probability is measured in terms of a ratio.
3. Probability can take values 0 to 1. It is expressed as percentage, proportion.
4. If one does not affect the occurrence of the other, then it said to be two events.
5. If A and B are two events which have no point in common, the events A and B are mutually exclusive.
6. Classical probability is also known as Laplace's probability.
7. Each outcome of a random experiment is called Primary event.
8. If A and B are two events, the probability of occurrence of either A or B is given as P (A U B).
9. The limiting relative frequency approach of probability is known as statistical probability.
10. The definition of statistical probability was originally given by Laplace.
11. If it is known that an event A has occurred, the probability of an event E given A is called conditional probability.
12. Probability by classical approach has many lecunae.
13. Classical probability is possible in case of equilikely outcomes.
14. An event consisting of those elements which are not in A is called as complementary event.
15. The probability of all possible outcomes of a random experiment is always equal to one.
16. The probability of the intersection of two mutually exclusive events is always zero.
17. The individual probabilities of occurrence of two events A and B are known, the probability of occurrence of both the events together will be decreased.
18. If two events A and B are such that A subset B and B subset A, the relation between P(A) and P(B) is P(A) = P(B).
19. If A is an event, the conditional probability of A given A is equal to one.
20. If A subset B, the probability, P (A/B) is equals to P(A) / P(B).
21. If B subset A, the probability, P (A/B) is equal to one.
22. If two events A and B are such that A subset B, the relation between the conditional probability P (A/C) and P(B/C) is P(A/C) < P(B/C).
23. For any two events A and B, P (A - B) is equal to P (A) - P (AB).
24. If an event B has occurred and it is known that P (B) = 1, the conditional probability P (A/B) is equal to P (A).
25. If A and B are any two mutually exclusive events, the P (A/A U B) is equal to P(A) / [P (A) + P (B)].
26. The idea of posteriori probabilities was introduced by Thomas Bayes.
27. In a city 60 percent read newspaper A, 40 % read newspaper B and 30 % read newspaper C, 20 % read A and B, 30 % read A and C, 10 % read B and C. Also 15 % read papers A, B, C. The percentage who does not read any these newspapers is 15%.
28. If a bag contains 4 white and 3 black balls. Two draws of 2 balls are successively made, the probability of getting 2 white balls at first draw and 2 black balls at second draw when the balls drawn at first draw were replaced is 2/49. If the balls are not replaced after the first draw, the probability of 2 black balls at second draw is 3/35.
29. In tossing 3 coins at a time, the probability of getting at most one head is 1/2.
30. There is 80 % chance that a problem will be solved by a statistics student and 60 % chance is there that the same problem will be solved by the mathematics student. The probability that at least the problem will be solved is 0.92.
31. The probability of two persons being borned on the same day is 1/7.
32. An urn contains 5 red, 4 white and 3 black balls. The probability of 3 balls being of different colors when the ball is replaced after each draw is equal to 2/144. The probability of 3 balls being drawn in the order red, white and black when the balls are not replaced after each draw, is equal to 1/22.
33. An urn A contains 5 white, and 3 black balls and B contains 4 white and 4 black balls. An urn is selected, and a ball is drawn from it, the probability, that the ball is white, is 5/16.
34. From a pack of 52 cards, two cards are drawn at random. The probability that one is an ace and the other is a king is 8/663.
35. Two dice are rolled by two players A and B. A throw 10, the probability that B throws more than A is 1/12.
36. The data reveals that 10 % patients die in a particular type of operation. A doctor performed 9 operations and all of them survived. Whether the 10th patient on being operated may survive or die.
37. There are two groups of students consisting of 4 boys and 2 girls: 3 boys and 1 girl. One student is selected from both the groups. The probability of one boy and one girl being selected is 5/12.
38. In a shooting competition, Mr. X can shoot at the bull's eye 4 times out of 5 shots and Mr. Y, 5 times out of six and Mr. Z, 3 times out of 4 shots. The probability that the target will be hit at least twice is 107/120.
39. There are two bags. One bag contains 4 red and 5 black balls and the other 5 red and 4 black balls. One ball is to be drawn from either of the two bags. The probability of drawing a black ball is 1/2.
40. 3 dice are rolled simultaneously. The probability of getting 12 spots is 25/216.
41. A bag contains 3 white and 5 red balls. 3 balls are drawn after shaking the bag. The odds against these balls being red is 5/28.
42. A bag contains 3 white, 1 black and 3 red balls. Two balls are drawn from the well shaked bag. The probability of both the balls being black is zero.
43. The chance of winning the race of the horse A in Durby is 1/5 and that of horse B is 1/. The probability that the race will be won by A or B is 11/30.
44. Four cards are drawn from a pack of 52 cards. The probability that out of 4 cards being 2 red and 2 black is 325/833.
45. For a 60-year-old person living up to the age of 70, it is 7:5 against him and for another 70-year-old person surviving up to the age of 80, it is 5:2 against him. The probability that one of them will survive for 10 years more is 49/84.
46. If 7:6 is in favour of A to survive 5 years more and 5:3 in favour of B to survive 5 years more, the probability that at least one of them will survive for 5 years more is 43/52.
47. The probability of throwing an odd sum with two fair dice is 1/2.
48. The probability that there is at least one spot in two rollings of a die is 11/36.
49. A group consists of 4 men, 3 women and 2 boys. 3 persons are selected at random. The probability that 2 men are selected is 5/14.
50. The probability that a leap year will have 53 sundays is 2/7.
51. If the chance of a A hiting a target is 3 times out of 4 and of B 4 times out of 5 and of C 5 times out of 6. The probability that the target will be hit in two hits is 47/120.
52. A consignment of 15 record players contains 4 defectives. The record players are selected at random one by one and examined. The ones examined are not put back. The probability that the 9th piece examined is the last defective one is 8/195.
53. The chance that doctors A will diagnose a disease X correctly is 60 %. The chance that a patient will die by this treatment after correct diagnosis is 40 %, and the chance of death by wrong diagnosis is 70%. A patient of doctor A, who had disease X, died. The probability that his disease was diagnosed correctly is 6/13.
54. An urn contains 5 yellow, 4 black and 3 white balls. 3 balls are drawn at random. The probability that no black ball is selected is 7/55.
55. A bag contains 3 white and 5 red balls. A game is played such that a ball is drawn, its color is noted and replaced with two additional balls of the same color. The selection is made 3 times. The probability that a white ball is selected at each trial is 7/64.
56. A can hit a target 2 times in 5 shots, B 3 times in 5 shots and C 4 times in 5 shots. They fire a volley. The probability that two shots hit is 58/125.
57. There are four coins in a bag. One of the coins has head on both sides. A coin is drawn at random and tossed five times and fell always with head upward. The probability that it is the coin with two head is 32/35.
58. One of the two events is certain to happen. The chance of one event is one-fifth of the other. The odds in favour of the other is 1:5.
59. One of the two events must happen; given that the chance of one is one-fourth of the other. The odd is favour of the other is 1:4.
60. A coin is tossed six times. The probability of obtaining heads and tails alternately is 1/32.
61. The odds in favour of the certain event are 5:4, and odds against another event are 4:3. The chance that at least one of them will happen is 47/63.
62. A and B start in a ring with ten other persons. If the arrangement of 12 persons is a random, the chance that there are exactly 3 persons between A and B is 2/11.
63. 3 houses were available in a locality for allotment. 3 persons applied for a house. The probablity that all the 3 persons applied for the same house is 1/9.
64. A speaks truth 4 times out of five and B speaks truth 3 times out of 4. They agree in the assertion that a white ball has been drawn from a bag containing 10 balls of different colours. The probability that a white ball was really drawn is 81/82.
65. The probability of drawing a white ball in the first draw and again a white ball in the second draw with replacement from a bag containing 6 white and 4 blue balls is 36/100.
66. A fair coin is tossed repeatedly unless a head is obtained. The probability that the coin has to be tossed at least 4 times is 1/4.
67. Out of 20 employees in a company, 5 are graduates. 3 employees are selected at random. The probability of all the 3 being graduates is 1/114.
68. A machine's part is produced by 3 factories A, B and C. Their proportional production is 25, 35 and 40 %, respectively. Also, the percentage defectives manufactured by 3 factories are 5,4 and 3, respectively. A part is taken at random and is found to be defective. The probability that the selected part belongs to factor B is 4/11.
69. A card is drawn from a well shuffled pack of 52 cards. A gambler bets that it is either a heart or an ace. What are odds against his winning this wet as 9:4.
70. An unbiased coin is tossed 4 times. The probability that the number of heads exceeds the number of tails is 5/16.