﻿ Probability Worksheets - Page 3 | Problems & Solutions
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Probability Worksheets
• Page 3
21.
If the event B depends on the event A, then P(B|A) = a. b. c. d.

#### Solution:

If the event B depends on the event A, then P(B|A) = P(A and B)P(A).
[Conditional Probability Formula.]

Correct answer : (1)
22.
A card is drawn from a well-shuffled deck of 52 cards and then a second card is drawn. Find the probability that the first card is a spade and the second card is a club if the first card is not replaced. a. $\frac{13}{51}$ b. $\frac{13}{102}$ c. $\frac{51}{52}$ d. $\frac{13}{204}$

#### Solution:

P(first spade card) = P(S) = 13 / 52 = 1 / 4

After the event of drawing a spade, the deck has 51 cards, of which 13 are clubs(C).

Therefore, P(C|S) = 13 / 51

Hence, P(S and C) = P(S) · P(C|S)
[Conditional Probability.]

= (14) · (1351)

= 13204

Correct answer : (4)
23.
Two dice were thrown and it is known that the numbers which come up were different. Find the probability that the sum of the two numbers was 4. a. $\frac{1}{15}$ b. $\frac{4}{15}$ c. $\frac{1}{30}$ d. $\frac{2}{15}$

#### Solution:

Let A be the event in which the two dice show different numbers and let B be the event in which the sum is 4. Then P(A) = 30 / 36

There are two outcomes (3, 1) and (1, 3) in which the sum is 4 and the numbers are different. Hence, P(B A) = 2 / 36

P(B|A) = P(B ∩ A)P(A)
[Conditional probability.]

= ( 236) ÷ ( 3036)

= 115

Correct answer : (1)
24.
A dice is thrown 6 times. If 'getting an even number' is a success, then what is the probability of exactly 5 successes? a. $\frac{5}{32}$ b. $\frac{3}{32}$ c. $\frac{1}{16}$ d. $\frac{1}{10}$

#### Solution:

In a single trail, probability p of success is p = P(getting an even number 2, 4 or 6)

= 36 = 12

Probability of failure, q = 1 - p

= 1 - 12 = 12

P(r) = (nr) (p)r(q)n - r
[Binomial distribution.]

P(exactly 5 successes) = (65) (12)5(12)1

= 332

Correct answer : (2)
25.
A coin is tossed 5 times. If 'getting a head' is considered as a success, find the probability of at least 3 successes. a. $\frac{1}{6}$ b. $\frac{1}{2}$ c. $\frac{1}{4}$ d. $\frac{1}{8}$

#### Solution:

In a single trial, probability p of success is given by p = P(getting a head) = 1 / 2

Probability of failure, q = 1 - p = 1 - 1 / 2 = 1 / 2

P(r) = (nr) (p)r(q)n - r
[Binomial distribution.]

P(r) = p(3) + p(4) + p(5)

= (53)(1 / 2)3(1 / 2)2 + (54)(1 / 2)4(1 / 2) + (1 / 2)5

= 1032 + 532 + 132

= 1632 = 12

Correct answer : (2)
26.
A box contains 20 cards labeled 1 through 20. One card is drawn from it at random. What is the probability of getting an odd number and the odds in favor of getting an odd number? a. $\frac{2}{5}$ and 1 : 1 b. $\frac{1}{10}$ and 1 : 2 c. $\frac{9}{10}$ and 1 : 2 d. $\frac{1}{2}$ and 1 : 1

#### Solution:

P (getting an odd number) = 10 / 20 = 1 / 2
[There are 10 odd numbers: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19.]

Odds in favor of getting an odd number are 10 : 10 or 1 : 1
[Ten of the outcomes are successes and 10 of them are failures.]

Correct answer : (4)
27.
Fifteen cards labeled A through O are placed in a bag. The bag is shaken and one card is drawn at random. Find the probability of not drawing a vowel. a. $\frac{4}{15}$ b. $\frac{4}{5}$ c. $\frac{1}{2}$ d. $\frac{11}{15}$

#### Solution:

P (not getting a vowel) = 11 / 15
[There are 11 ways of not getting a vowel: B, C, D, F, G, H, J, K, L, M, N.]

Correct answer : (4)
28.
If a card is drawn from a bridge deck, what is the probability of getting a king? a. $\frac{1}{2}$ b. $\frac{1}{13}$ c. $\frac{1}{52}$ d. $\frac{1}{4}$

#### Solution:

Let E be the event of getting a king from the 4 in the deck.

Number of outcomes in E = 4C1

= 4! / 1!(4-3)!
[nCr = n! / r!(n-r)! .]

= 4×3! / 1!3! = 4

Let S be the sample space consists of all cards.

Number of outcomes in S = 52C1

= 52! / 1!(52-1)! = 52
[nCr = n! / r!(n-r)! .]

P(E) = Number of outcomes in the eventNumber of outcomes in the sample space

= 4 / 52= 1 / 13

Correct answer : (2)
29.
A box contains 20 cards labeled 1 through 20 on them. One card is drawn from it at random. Find the probability of getting an even number. a. $\frac{9}{20}$ b. $\frac{1}{2}$ c. $\frac{1}{20}$ d. $\frac{1}{4}$

#### Solution:

P (getting an even number) = 10 / 20 = 1 / 2
[There are 10 even numbers: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20.]

Correct answer : (2)
30.
A box contains 20 cards labeled 1 through 20. One card is drawn from it at random. Find the probability of getting a prime number. a. $\frac{1}{2}$ b. $\frac{9}{20}$ c. $\frac{1}{5}$ d. $\frac{2}{5}$

#### Solution:

P (getting a prime number) = 8 / 20 = 2 / 5
[There are 8 prime numbers: 2, 3, 5, 7, 11, 13, 17, 19.]

Correct answer : (4)

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