Probability - Aptitude MCQ Questions and Solutions with Explanations

Here, S = {1, 2, 3, 4, ...., 19, 20}
Let E = event of getting a multiple of 3 or 5
= {3, 6 , 9, 12, 15, 18, 5, 10, 20}
$$\therefore P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}} = \frac{9}{{20}}$$

Total number of balls
= (2 + 3 + 2)
= 7
Let S be the sample space
Then, n(S) = Number of ways of drawing 2 balls out of 7
$$\eqalign{ & {\text{n}}\left( {\text{S}} \right) = {}^7{C_2} \cr & {\text{n}}\left( {\text{S}} \right) = \frac{{\left( {7 \times 6} \right)}}{{\left( {2 \times 1} \right)}} \cr & {\text{n}}\left( {\text{S}} \right) = 21 \cr} $$
Let E = Event of 2 balls, none of which is blue
∴ n(E) = Number of ways of drawing 2 balls out of (2 + 3) balls
$$\eqalign{ & {\text{n}}\left( {\text{E}} \right)\, = {}^5{C_2} \cr & {\text{n}}\left( {\text{E}} \right) = \frac{{\left( {5 \times 4} \right)}}{{\left( {2 \times 1} \right)}} \cr & {\text{n}}\left( {\text{E}} \right) = 10 \cr & \therefore {\text{P}}\left( {\text{E}} \right) = \frac{{{\text{n}}\left( {\text{E}} \right)}}{{{\text{n}}\left( {\text{S}} \right)}} = \frac{{10}}{{21}} \cr} $$

Total number of balls
= (8 + 7 + 6)
= 21
Let E = event that the ball drawn is neither red nor green
= event that the ball drawn is blue
$$\eqalign{ & \therefore n(E) = 7 \cr & \therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{7}{{21}} = \frac{1}{3} \cr} $$

In two throws of a dice, n(S) = (6 x 6) = 36
Let E = event of getting a sum
= {(3, 6), (4, 5), (5, 4), (6, 3)}
$$\eqalign{ & \therefore P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}} \cr & = \frac{4}{{36}}\cr & = \frac{1}{9} \cr} $$

Getting at most Two heads means 0 to 2 but not more than 2
Here S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}
Let E = event of getting at most two heads
Then E = {TTT, TTH, THT, HTT, THH, HTH, HHT}
$$\therefore P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}} = \frac{7}{8}$$

In a simultaneous throw of two dice, we have n(S) = (6 x 6) = 36

Then, E = {(1, 2), (1, 4), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 2), (3, 4), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 2), (5, 4), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
$$\eqalign{ & \therefore n\left( E \right) = 27 \cr & \therefore P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}} = \frac{{27}}{{36}} = \frac{3}{4} \cr} $$

Let S be the sample space and E be the event of selecting 1 girl and 2 boys
Then, n(S) = Number ways of selecting 3 students out of 25
$$\eqalign{ & = {}^{25}{C_3} \cr & = \frac{{ {25 \times 24 \times 23} }}{{ {3 \times 2 \times 1} }} \cr & = 2300 \cr & n\left( E \right) = {^{10}{C_1}{ \times ^{15}}{C_2}} \cr & = {10 \times \frac{{ {15 \times 14} }}{{ {2 \times 1} }}} \cr & = 1050 \cr & \therefore P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}} = \frac{{1050}}{{2300}} = \frac{{21}}{{46}} \cr} $$

$$\eqalign{ & P\left( {{\text{getting}}\,{\text{a}}\,{\text{prize}}} \right) \cr & = \frac{{10}}{{10 + 25}} \cr & = \frac{{10}}{{35}} \cr & = \frac{2}{7} \cr} $$

Let S be the sample space
$$\eqalign{ & {\text{Then}},n\left( S \right) = {}^{52}{C_2} \cr & = \frac{{ {52 \times 51} }}{{\left( {2 \times 1} \right)}} \cr & = 1326 \cr} $$
Let E = event of getting 2 kings out of 4
$$\eqalign{ & \therefore n\left( E \right) = {}^4{C_2} = \frac{{ {4 \times 3} }}{{ {2 \times 1} }} = 6 \cr & \therefore P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}} \cr & = \frac{6}{{1326}} \cr & = \frac{1}{{221}} \cr} $$

Clearly, n(S) = (6 x 6) = 36
Let E = Event that the sum is a prime number.Then
E = {(1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4, 3), (5, 2), (5, 6), (6, 1), (6, 5)}
$$\eqalign{ & \therefore n\left( E \right) = 15 \cr & \therefore P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}} \cr & = \frac{{15}}{{36}} \cr & = \frac{5}{{12}} \cr} $$

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Alternate solution

Understanding the Question:
We're tossing two dice. Each die has six sides (1, 2, 3, 4, 5, 6).
We want to find the probability that the sum of the numbers shown on both dice is a prime number.
A prime number is a number greater than 1 that is only divisible by 1 and itself (e.g., 2, 3, 5, 7, 11...).

Finding the Possible Outcomes:
First, let's figure out all the possible sums we can get when we add the numbers on two dice. The minimum sum is 2 (1+1) and the maximum is 12 (6+6).

Identifying Prime Number Sums:
Now, let's list the sums that are prime numbers: 2, 3, 5, 7, 11.

Counting Favorable Outcomes:
Let's count how many ways we can get each of these prime sums:
* 2: Only one way (1+1)
* 3: Two ways (1+2, 2+1)
* 5: Four ways (1+4, 2+3, 3+2, 4+1)
* 7: Six ways (1+6, 2+5, 3+4, 4+3, 5+2, 6+1)
* 11: Two ways (5+6, 6+5)
Total favorable outcomes (prime number sums): 1 + 2 + 4 + 6 + 2 = 15

Calculating Total Possible Outcomes:
The total number of possible outcomes when tossing two dice is 6 (outcomes for the first die) * 6 (outcomes for the second die) = 36

Calculating Probability:
Probability is calculated as (Favorable Outcomes) / (Total Possible Outcomes).
So, the probability of getting a prime number sum is 15/36. This simplifies to 5/12.

Therefore, the correct option is B: 5/12