Probability- Aptitude Questions and Answers

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A card is drawn from a well shuffled pack of cards. What is the probability that it is a red card or a face card?

A)
19/26

B)
9/13

C)
8/13

D)
37/52


Correct Answer :

8/13


Explanation :

There are a total of 52 cards.
Now, there are 26 red cards and 12 face cards, 6 of which are red (double counted)

So, the probability of having a card's red card or face card
P(Red u King)=P(Red)+P(face)-P(Red n face)
=26/52 + 12/52 - 6/52
=32/52
=8/13

A box of balls contains 120 red and 80 blue one. Two balls are taken at random. What is the probability that one is red and the other is blue ?

A)
96/199

B)
16/33

C)
24/100

D)
8/33


Correct Answer :

96/199


Explanation :

Total number of ball are 200.
Probability of getting one red and one blue => (120C1 * 80C1 ) / 200C2
=>(120* 80) / (200 *199 / 2 )
=>96/199

A box contains 20 white, 30 black, 40 blue and 30 red balls. Compute the probability that one of the balls extracted at random from the box turns out to be white, black or red. ?

A)
3/4

B)
3/5

C)
1/3

D)
2/3


Correct Answer :

2/3


Explanation :

Probability that the extracted ball is blue = 40/120 => 1/3
So, Probability that the extracted ball is not blue = 1- (1/3) => 2/3

If seven coins are tossed, what is the probability of obtaining at least 2 heads ?

A)
12/16

B)
15/16

C)
15/18

D)
15/20


Correct Answer :

15/16


Explanation :

Probability of getting no head =7C0 / 27 => 1/128
Probability of getting one head =7C1 / 27=> 7/128

Probability of getting at least 2 heads= 1 - 1/128 - 7/128 => 120/128 =>15/16

If one rolls a fair-sided die twice, what is the probability that the die will land on the same number on both the occasions ?

A)
1/12

B)
1/6

C)
1/24

D)
1/8


Correct Answer :

1/6


Explanation :

n(S), Total ways = 6∗6= 36
n(E), Event of getting same number on both occasions
=> (1, 1), (2, 2), (3, 3), (4, 4),(5, 5), (6, 6) = 6

Required probability = n(E)/n(S) = 6 / 36 = 1/6

A die is thrown once. Find the probability of getting an even number?

A)
3/4

B)
1/4

C)
1/2

D)
3/2


Correct Answer :

1/2


Explanation :

Sample space is, S : {1,2,3,4,5,6} =6

Favorable outcomes,even numbers E : { 2,4,6 } = 3

P(getting an event number)=n(E)/n(S)=3/6=1/2

A card is drawn from a well-shuffled deck of 52 playing cards. Find the probability that it is not an ace?

A)
1/13

B)
4/13

C)
3/13

D)
12/13


Correct Answer :

12/13


Explanation :

total number of cards=52
total number of aces=4
The probability P(getting an ace)=n(E)/n(S)=4/52=1/13

The probability P(not getting an ace)=1 - P(getting an ace)
=1 - (1/13)
=12/13

A bag contains 24 eggs out of which 8 are rotten. The remaining eggs are not rotten eggs. The two eggs are selected at random. What is the probability that one of the eggs is rotten?

A)

11/23


B)

17/23


C)

13/23


D)

19/23



Correct Answer :

13/23


Explanation :

Number of total eggs = 24
Number of rotten egss = 8
Remaining eggs = 24 − 8 = 16

Required probability = (16C1 * 8C1 + 16C0 * 8C2 ) / 24C2
= 16 * 8 + (8 * 7 / 2) / (24 * 23 / 2 )
= 16 * 8 + 4 * 7 / 12 * 23
= 128 + 28 / 12 * 23
= 156 / 12 * 23
= 13 / 23

From a well-shuffled deck of 52 cards, one card is drawn at random. Find the probability of getting a face card?

A)

1/13


B)

2/13


C)

3/13


D)

4/13



Correct Answer :

3/13


Explanation :

Sample space is, S : 52 cards
Favorable outcomes, E : 12 face cards (4 jack, 4 queen and 4 king)

Probability, P(face card) = Favorable outcomes/Sample space
= 12/52
= 3/13

What is the probability that a number selected from the numbers 1, 2, 3,…….,25 is a prime number, when each of the given numbers is equally likely to be selected?

A)

8/25


B)

9/25


C)

1/25


D)

2/25



Correct Answer :

9/25


Explanation :

Let S be the sample space.
Then, n(S) = numbers between 1 to 25 = 25

Let E be the event space= event that the number selected is a prime number = {2, 3, 5, 7, 11, 13, 17, 19, 23}
n(E) = 9

Probability of occurrence of event, P(E) = n(E)/n(S) = 9/25

The probability of Sita, Gita and Mita passing a test is 60 %, 40 % and 20 % respectively. What is the probability that at Sita and Gita will pass the test and Mita will not?

A)
38.4%

B)
60%

C)
4.8%

D)
19.2%


Correct Answer :

19.2%


Explanation :

Probability of Mita failing = 100 - Probability of Mita passing
= 100 - 20
= 80

The probability of Sita passing (60%), Gita passing (40%), and Mita failing (80%), combine with the formula
P(A and B and C) = P(A)* P(B) * P(C),
so 0.6 * 0.4 * 0.8 = .192
giving a 19.2% chance that Sita and Gita will pass and Mita will not pass.

An unbiased die is tossed.Find the probability of getting a multiple of 3?

A)

1/2


B)
1/3

C)
1/4

D)

2/3



Correct Answer :

1/3


Explanation :

Here we have sample space S={1,2,3,4,5,6}
n(S) =6

Let E be the event of getting a multiple of 3 then E={3,6}
n(E) =2

P(E) =n(E)/n(S)
P(E) =2/6 =1/3

In a simultaneous throw of a pair of dice,find the probability of getting a total more than 7?

A)

1/2


B)
5/12

C)
7/15

D)

3/12



Correct Answer :

5/12


Explanation :

Here sample space n(S) =6*6 =36.
Let E be the event of getting a total more than 7 ={(1,6),(2,5),(3,4),(4,3)(5,2),(6,1)(2,6),(3,5),(4,4),(5,3),(6,2),(4,5),(5,4),(5,5),(4,6),(6,4)}
n(E) =15
P(E) = n(E)/n(S) = 15/36 = 5/12

A bag contains 6 white and 4 black balls .Two balls are drawn at random .Find the probability that they are of the same colour?

A)

1/15


B)
2/15

C)
4/15

D)

7/15



Correct Answer :

7/15


Explanation :

Let S be the sample space.
Then Number of ways for drawing two balls out of 6 white and 4 red balls, n(S) = 10C2
=10!/(8!*2!)
= 45

Let E =event of getting both balls of the same colour.
Then number of ways of drawing ( 2 balls out of 6) or (2 balls out of 4), n(E) =
= 6C2 +4C2
= 6!/(4!*2!) + 4!/(2! *2!)
= 6*5/2*1 + 4*3/2*1
= 15+6
= 21

Required probability, P(E) =n(E)/n(S) =21/45 =7/15

Two dice are thrown together .What is the probability that the sum of the number on the two faces is divisible by 4 or 6?

A)

7/18


B)
14/35

C)
8/18

D)

7/35



Correct Answer :

7/18


Explanation :

n(S) = 6*6 =36
E be the event for getting the sum of the number on the two faces is divisible by 4 or 6.
E={(1,3)(1,5)(2,4?)(2,2)(3,5)(3,3)(2,6)(3,1)(4,2)(4,4)(5,1)(5,3)(6,2)(6,6)}
n(E) =14

Hence P(E) =n(E)/n(S)
= 14/36
= 7/18

Two cards are drawn at random from a pack of 52 cards What is the probability that either both are black or both are queens?

A)

52/221


B)
55/190

C)
55/221

D)

19/221



Correct Answer :

55/221


Explanation :

total number of ways for choosing 2 cards from 52 cards is
= 52C2 = 52!/(50!*2!) = 1326

Let A= event of getting both black cards
and B= event of getting both queens
A ∩ B=Event of getting queens of black cards

We have 26 black cards from that we have to choose 2 cards.
n(A)=26C2=26!/(24!*2!) = 26*25/2*1=325

In 52 cards we have 4 queens from that we have to choose 2 cards.
n(B) = 4C2 = 4!/(2!* 2!) = 4*3/2*1 =6

from both case, we have to choose 2 black queens.
n(A∩B) =2C2 =1

P(A) = n(A) /n(S) =325/1326
P(B) = n(B)/n(S) = 6/1326
P(A∩B) = n(A∩B)/n(S) = 1/1326
P(AUB) = P(A) +P(B) -P(A∩B)
= 325/1326 + 6/1326 -1/1326
= 330/1326
= 55/221

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