Probability Practice Test Questions – Aptitude -- Free Practice

Question 1

Two cards are drawn at random from a pack of 52 cards. The probability of getting either both red cards or both King cards is:

A

0.2488✓✗

B

0.4288✓✗

C

0.8248✓✗

D

0.8428✓✗

Explanation

P(both red) = C(26,2)/C(52,2) = 325/1326
P(both Kings) = C(4,2)/C(52,2) = 6/1326
P(both red and both Kings) = C(2,2)/C(52,2) = 1/1326 (the two red Kings)
P(both red or both Kings) = 325/1326 + 6/1326 - 1/1326 = 330/1326 ≈ 0.2488

Question 2

The probability of success of three students in the CA Foundation examination are 1/5, 1/4 and 1/3 respectively. Find the probability that at least two students will get success.

A

25✓✗

B

34✓✗

C

16✓✗

D

15✓✗

Explanation

P(all three succeed) = 1/5 × 1/4 × 1/3 = 1/60
P(exactly two succeed) = (1/5 × 1/4 × 2/3) + (1/5 × 3/4 × 1/3) + (4/5 × 1/4 × 1/3) = 2/60 + 3/60 + 4/60 = 9/60
P(at least two succeed) = 9/60 + 1/60 = 10/60 = 1/6

Question 3

If a coin is tossed 5 times, then the probability of getting Tail and Head occurring alternatively is:

A

18✓✗

B

116✓✗

C

132✓✗

D

164✓✗

Explanation

There are exactly 2 alternating sequences of 5 tosses: THTHT and HTHTH
P = 2 × (1/2)⁵ = 2/32 = 116

Question 4

When 2 dice are thrown simultaneously, the probability of getting at least one 5 is:

A

1136✓✗

B

536✓✗

C

815✓✗

D

17✓✗

Explanation

P(no 5 on either die) = (5/6)² = 25/36
P(at least one 5) = 1 - 25/36 = 1136

Question 5

The probability that a leap year has 53 Wednesdays is:

A

27✓✗

B

35✓✗

C

17✓✗

D

23✓✗

Explanation

A leap year has 366 days = 52 weeks + 2 extra days
These 2 extra days can be any of 7 equally likely combinations, of which 2 include a Wednesday (Tue-Wed, Wed-Thu)
P = 27

Question 6

Find the probability of success for the binomial distribution satisfying the relation 4P(x = 4) = P(x = 2), with parameter n = 6.

A

13✓✗

B

12✓✗

C

15✓✗

D

18✓✗

Explanation

4 × C(6,4) p⁴q² = C(6,2) p²q⁴
Since C(6,4) = C(6,2) = 15, this gives 4p² = q², so q = 2p
Since p + q = 1: p + 2p = 1 ⇒ p = 13

Question 7

In a business venture, a man can make a profit of Rs. 50,000 or incur a loss of Rs. 20,000. The probabilities of making a profit or incurring a loss, from past experience, are known to be 0.75 and 0.25 respectively. What is his expected profit?

A

Rs. 33,500✓✗

B

Rs. 34,500✓✗

C

Rs. 35,500✓✗

D

Rs. 32,500✓✗

Explanation

Expected profit = (50000 × 0.75) - (20000 × 0.25) = 37,500 - 5,000 = Rs. 32,500

Question 8

Rupesh is known to hit a target in 5 out of 9 shots, whereas David is known to hit the same target in 6 out of 11 shots. What is the probability that the target would be hit once they both try?

A

7999✓✗

B

1013✓✗

C

1426✓✗

D

1318✓✗

Explanation

P(neither hits) = 49 × 511 = 2099
P(at least one hits) = 1 - 2099 = 7999

Question 9

The probability that an Accountant's job applicant has a B.Com. degree is 0.85, that he is a CA is 0.30, and that he is both B.Com. and CA is 0.25. What is the probability that an applicant is B.Com. or CA?

A

0.25✓✗

B

0.30✓✗

C

0.10✓✗

D

0.90✓✗

Explanation

P(B.Com. or CA) = P(B.Com.) + P(CA) - P(both) = 0.85 + 0.30 - 0.25 = 0.90

Question 10

Find the probability that a four-digit number comprising the digits 2, 5, 6 and 7 (each used exactly once) would be divisible by 4.

A

14✓✗

B

13✓✗

C

12✓✗

D

1✓✗

Explanation

Total 4-digit numbers = 4! = 24
A number is divisible by 4 if its last two digits form a number divisible by 4. Checking all pairs from {2, 5, 6, 7}, the valid last-two-digit combinations are 52, 56, 72, 76 (4 combinations), each with 2 arrangements of the remaining digits = 8 favourable numbers
P = 824 = 13

Question 11

A manufacturer, who produces medicine bottles, finds that 0.1% of the bottles are defective. The bottles are packed in boxes containing 500 bottles. A drug manufacturer buys 100 boxes from the producer of bottles. Using the Poisson distribution, find how many boxes will contain at least two defectives. [Given: e^-0.5 = 0.6065]

A

7✓✗

B

13✓✗

C

9✓✗

D

11✓✗

Explanation

λ = 500 × 0.001 = 0.5 per box
P(X≥2) = 1 - P(0) - P(1) = 1 - e^-0.5 - 0.5e^-0.5 = 1 - 1.5e^-0.5
= 1 - 1.5 × 0.6065 = 1 - 0.90975 = 0.09025
Expected number of boxes (out of 100) = 100 × 0.09025 ≈ 9

Question 12

Two unbiased dice are thrown. The expected value of the sum of numbers on the upper side is:

A

3.5✓✗

B

7✓✗

C

12✓✗

D

6✓✗

Explanation

E(sum) = E(die1) + E(die2) = 3.5 + 3.5 = 7

Question 13

Probability of getting exactly one head when two unbiased coins are tossed simultaneously is:

A

0.25✓✗

B

0.50✓✗

C

0.20✓✗

D

0.75✓✗

Explanation

Sample space for two coins: {HH, HT, TH, TT} -- 4 equally likely outcomes.
Exactly one head: HT, TH -- 2 outcomes.
Probability = 2/4 = 0.50

Question 14

There are two boxes containing 5 white and 6 blue balls, and 3 white and 7 blue balls respectively. If one of the boxes is selected at random and a ball is drawn from it, then the probability that the ball is blue is:

A

115/227✓✗

B

83/250✓✗

C

137/220✓✗

D

127/250✓✗

Explanation

P(blue) = P(box1) × P(blue|box1) + P(box2) × P(blue|box2)
= 12 × 611 + 12 × 710
Using 220ths: 60220 + 77220 = 137220

Question 15

A box contains 5 white and 7 black balls. Two successive draws of 3 balls are made (i) with replacement (ii) without replacement. The probability that the first draw would produce white balls and the second draw would produce black balls are respectively:

A

6/321 and 3/926✓✗

B

1/20 and 1/30✓✗

C

35/144 and 35/108✓✗

D

7/968 and 5/264✓✗

Explanation

With replacement: P(3 white) = C(5,3)/C(12,3) = 10/220 = 1/22. Since the pool resets, P(3 black) = C(7,3)/C(12,3) = 35/220 = 7/44.
Combined (independent) = 122 × 744 = 7968
Without replacement: after removing 3 white, 9 balls remain (2 white + 7 black). P(3 black) = C(7,3)/C(9,3) = 35/84 = 5/12.
Combined = 122 × 512 = 5264

Question 16

A number is selected at random from the set {1, 2, ....., 99}. The probability that it is divisible by 9 or 11 is:

A

19/100✓✗

B

19/99✓✗

C

10/100✓✗

D

10/99✓✗

Explanation

Multiples of 9 from 1-99: 11. Multiples of 11: 9. Multiples of both (i.e. 99): 1.
Divisible by 9 or 11 = 11 + 9 - 1 = 19
Probability = 1999

Question 17

A random variable X follows a Binomial Distribution with E(X) = 2 and V(X) = 1.2. The value of n is:

A

8✓✗

B

2✓✗

C

5✓✗

D

None of these✓✗

Explanation

E(X) = np = 2. V(X) = npq = 1.2
q = npqnp = 1.22 = 0.6, so p = 0.4
n = 20.4 = 5

Question 18

The mean of a Poisson distribution is 4. The probability of exactly two successes is:

A

8/e⁴✓✗

B

4/e⁴✓✗

C

16/e⁴✓✗

D

8/e²✓✗

Explanation

For Poisson with λ=4: P(X=2) = e^-4 × 4²2! = e^-4 × 162 = 8e^-4 = 8e⁴

Question 19

A die is thrown 100 times. If getting an even number is considered a success, find the variance of the number of successes.

A

50✓✗

B

25✓✗

C

10✓✗

D

100✓✗

Explanation

P(success) = P(even number) = 3/6 = 0.5. n = 100, p = 0.5, q = 0.5
Variance = npq = 100 × 0.5 × 0.5 = 25

Question 20

The chance of getting a sum of 10 in a single throw of two dice is:

A

10/36✓✗

B

1/12✓✗

C

1/36✓✗

D

None of these✓✗

Explanation

Ways to get a sum of 10 with two dice: (4,6), (5,5), (6,4) -- 3 ways, out of 36 total outcomes.
Probability = 336 = 112

Question 21

Find the expected value of the following distribution:

x	-20	-10	30	75	80
P(x)	320	15	12	110	120

A

20.5✓✗

B

21.5✓✗

C

22.5✓✗

D

24.5✓✗

Explanation

E(X) = (-20)(3/20) + (-10)(1/5) + 30(1/2) + 75(1/10) + 80(1/20)
= -3 + (-2) + 15 + 7.5 + 4 = 21.5

Question 22

If P(A) = 4/9, then the odds against the event 'A' are:

A

4:9✓✗

B

4:5✓✗

C

5:4✓✗

D

4:14✓✗

Explanation

P(A) = 4/9, so P(not A) = 5/9
Odds against A = P(not A) : P(A) = 5/9 : 4/9 = 5:4

Question 23

Find the variance of a binomial distribution with n = 10, p = 0.3

A

2.1✓✗

B

3✓✗

C

7✓✗

D

None of these✓✗

Explanation

Variance = npq = 10 × 0.3 × 0.7 = 2.1

Question 24

One card is drawn from a pack of 52, what is the probability that it is a king or a queen?

A

11/13✓✗

B

2/13✓✗

C

1/13✓✗

D

None of these✓✗

Explanation

There are 4 kings and 4 queens = 8 favourable cards out of 52.
Probability = 852 = 213

Question 25

A bag contains 8 red and 5 white balls. Two successive draws of 3 balls are made without replacement. The probability that the first draw will produce 3 white balls and the second 3 red balls is:

A

6/255✓✗

B

5/548✓✗

C

7/429✓✗

D

3/233✓✗

Explanation

P(3 white first) = C(5,3)C(13,3) = 10286
P(3 red second, from remaining 10 balls) = C(8,3)C(10,3) = 56120
P(both) = 10286 × 56120 = 56034320 = 7429