Probability- Aptitude Questions and Answers

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

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

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

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

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

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

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

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

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

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

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.

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

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

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

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

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