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.
