Probability: Aptitude Questions with Answers
Q. 1 Rahim is practicing archery. He is shooting at a target that his coach has set up for him. The chance that he hits the target with each arrow is 0.7. What is the estimated probability that he will successfully hit the target exactly three times out of his first five shots?
A) 0.1
B) 0.3
C) 0.85
D) 0.5
Check Solution
Ans: B
087 Given:
Probability of hitting the target is 0.7
Probability of Not hitting the target is 0.3
Now,probability that he will hit the target only three times in his first five attempts, this means, out of 5 attempts he will hit the target 3 times and miss the target 2 times ==> which can be represented as = 0.7 * 0.7 * 0.7 * 0.3 * 0.3
Now the three successful and unsuccessful attempts can happen in any sequence and which needs to be accounted for above.
So one of the sequence may be – Success,Failure,Success,Success,Failure
Another sequence may be – Failure, Success,Failure,Success,Success
So to account the different sequences we can considers that there are five places where three Success needs to be placed and this can be represented by 5C3 = 5!/3!*2! = 10
Therefore,
the approximate probability that he will hit the target only three times in his first five attempts
= 0.7 * 0.7 * 0.7 * 0.3 * 0.3 * 10
= 0.3087
Q. 2 You’re at a game night rolling two standard six-sided dice. You’re feeling lucky and decide to bet on the total sum of the dice being exactly 8. What are the odds that your roll will make you a winner?
A) $( \frac{5}{36} ) $
B) $( \frac{6}{36} ) $
C) $( \frac{7}{36} )$
D) $( \frac{4}{36} )$
Check Solution
Ans: A) $( \frac{5}{36} )$
Possible outcomes for a sum of 8 are ( (2,6), (3,5), (4,4), (5,3), (6,2) ).
Number of favorable outcomes = 5.
Total outcomes = $( 6 \times 6 = 36 )$.
Probability = $( \frac{5}{36} )$.
Q. 3 You’re holding a freshly shuffled deck of 52 cards, ready for some card magic. With a dramatic flair, you draw a single card. What’s the probability that your card is either a queen or a diamond from the deck?
A) $( \frac{17}{52} ) $
B) $( \frac{4}{13} )$
C) $( \frac{16}{52} )$
D) $( \frac{15}{52} )$
Check Solution
Ans: B) $( \frac{4}{13} )$
There are 4 queens and 13 diamonds in the deck.
One queen is also a diamond.
Favorable outcomes = ( 4 + 13 – 1 = 16 ).
Probability = $( \frac{16}{52} = \frac{4}{13} )$.
Q. 4 Imagine flipping a shiny, fair coin twice, hoping to see at least one head appear. What’s the probability that your coin tosses will deliver at least one head?
A) $( \frac{3}{4} )$
B) $( \frac{1}{4} )$
C) $( \frac{1}{2} )$
D) $( \frac{7}{8} )$
Check Solution
Ans: A) $( \frac{3}{4} )$
Probability of no head (both tails) = $( \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} )$.
Probability of at least one head = $( 1 – \frac{1}{4} = \frac{3}{4} )$.
Q. 5 In a box of 10 light bulbs, 3 of them have unfortunately turned out defective. If you randomly select 2 bulbs from the box without peeking, what’s the likelihood that both bulbs you pick are the faulty ones?
A) $( \frac{1}{15} )$
B) $( \frac{1}{30} )$
C) $( \frac{1}{10} )$
D) $( \frac{1}{20} )$
Check Solution
Ans: A) $( \frac{1}{15} )$
Total ways to choose 2 bulbs = $( \binom{10}{2} = 45 )$.
Ways to choose 2 defective bulbs = $( \binom{3}{2} = 3 )$.
Probability = $( \frac{3}{45} = \frac{1}{15} )$.
Q. 6 You’re tasked with forming a committee of 3 members from a pool of 5 men and 3 women. To ensure diversity, you want at least one woman on the committee. What’s the probability that your committee will include at least one woman?
A) $( \frac{35}{56} )$
B) $( \frac{21}{56} $
C) $( \frac{41}{56} $
D) $( \frac{47}{56} )$
Check Solution
Ans: C) $( \frac{41}{56} )$
Total ways = $( \binom{8}{3} = 56 )$.
Ways to select only men = $( \binom{5}{3} = 10 )$.
Ways with at least one woman = $( 56 – 10 = 46 )$.
Probability = $( \frac{46}{56} = \frac{41}{56} )$.
Q. 7 You toss three fair coins into the air, watching them spin and fall with a mix of anticipation and excitement. What is the probability that exactly two of those coins land showing heads?
A) $( \frac{1}{8} )$
B) $( \frac{3}{8} )$
C) $( \frac{1}{4} )$
D) $( \frac{1}{2} )$
Check Solution
Ans: B) $( \frac{3}{8} )$
Total outcomes = $( 2^3 = 8 )$.
Favorable outcomes = $( { HHT, HTH, THH } = 3 )$.
Probability = $( \frac{3}{8} )$.
Understanding common probability scenarios can help solve these questions easily.
Q. 8 A street cricket team of 3 is to be formed from 5 batsmen and 3 bowlers. You are hoping to see at least one bowler in the team, who can take some wickets. What is the probability that your wish will come true?
A) $\frac{35}{56}$
B) $\frac{21}{56}$
C) $\frac{41}{56}$
D) $\frac{47}{56}$
Check Solution
Ans: C) $\frac{41}{56}$
Total ways = $\binom{8}{3} = 56$
Ways to select only men = $\binom{5}{3} = 10$
Ways with at least one woman = $56−10=46$
Probability = $\frac{46}{56} = \frac{41}{56}$
Q. 9 The probability of rain in Surat is calculated as $\frac{2}{5}$ by weather department for tomorrow, what is the probability that it doesn’t rain in Surat tomorrow?
A) $\frac{1}{5}$
B) $\frac{3}{5}$
C) $\frac{4}{5}$
D) $\frac{2}{5}$
Check Solution
Ans: B) $\frac{3}{5}$
Probability of not happening = $1 – \frac{2}{5} = \frac{3}{5}$
Q. 10 A bag contains 4 white and 5 black balls. If you randomly draw two balls without looking, what is the probability that both of the balls you select turn out to be black?
A) $\frac{2}{9}$
B) $\frac{1}{3}$
C) $\frac{5}{18}$
D) $\frac{10}{36}$
Check Solution
Ans: C) $\frac{5}{18}$
Total ways to choose 2 balls = $\binom{9}{2} = 36$
Ways to choose 2 black balls = $\binom{5}{2} = 10$
Probability = $\frac{10}{36} = \frac{5}{18}$
Q. 11 Eyana is choosing toppings for a pizza and he can select three different ones from the following options: pepperoni, sausage, black olives, green peppers, and mushrooms. What are the total number of unique topping combinations he can create?
A) 5
B) 10
C) 15
D) 30
Check Solution
Ans: B
There are 10 different combinations of toppings available.
To solve this, we can use the combination formula:
$nCr = \dfrac{n!}{(n-r)!r!}$
where $n$ is the total number of items, $r$ is the number of items being chosen, and $!$ is the factorial function.
In this case, we have $n=5$ items (pepperoni, sausage, black olives, green peppers, and mushrooms) and we are choosing $r=3$ items.
Plugging in, we get:
$C_{5,3} = \dfrac{5!}{(5-3)!3!} = \dfrac{5 \times 4 \times 3}{2 \times 1} = 10$
Therefore, there are 10 different combinations of toppings available.
Q. 12 A gum ball machine contains 24 gum balls, split evenly with 12 white and 12 black. They are dispensed randomly. If the first three gum balls dispensed are black, what is the probability that the following two gum balls dispensed will be black as well?
A) 6/35
B) 4/15
C) 1/3
D) 3/7
Check Solution
Ans: A
Given data : 24 gum balls – 12 white and 12 black
3 black gum balls have already been chosen
To find : Probability that next 2 gum balls are also black in color
Denominator(2 gum balls chosen from 21 remaining balls) $21c2 = \frac{21*20}{2} = 210$
Numerator(Balls picked are black) $9c2 = \frac{9*8}{2} = 36$
Probability that next 2 gum balls are also black in color(of remaining 9 black gum balls) = $\frac{36}{210} = \frac{6}{35}$
Test your skills with these fun problems:
1. Squid Game Marble Challenge
2. Analyzing SRK’s Film Hit Probability
3. Rihanna Raffle Probability Puzzle