Probability - Aptitude MCQ Questions and Solutions with Explanations

$$\eqalign{ & P\left( A \right) = \frac{2}{5} \cr & P\left( B \right) = \frac{4}{7} \cr} $$
E = {A and B both get selected}
$$\eqalign{ & P\left( E \right) = {\text{ }}P\left( A \right) \times P\left( B \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{2}{5} \times \frac{4}{7} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{8}{{35}} \cr} $$

A dice has 6 faces
So there are 6 possible outcomes
Dice is rolled once and then again
So total possibilities = 6 × 6 = 36
The sum should be 8 of the 2 throws
So which combination of numbers from 1 to 6 will yield us a sum of 8
They are - (2, 6); (6, 2); (3, 5); (5, 3); (4, 4);
So there are total 5 possibilities where addition is 8.
But only 1 possibility where first throw of dice is 4.
So, Probability for first throw to be 4 and sum to be 8 = $$\frac{{1}}{{36}}$$

In a simultaneous throw of two dice, we have n(S) = 6 x 6 = 36
Let E = event of getting two numbers whose sum is odd.
Then E = {(1,2), (1,4), (1,6), (2,1), (2,3), (2,5), (3,2), (3,4), (3,6), (4,1), (4,3), (4,5), (5,2), (5,4), (5,6), (6,1), (6,3), (6,5)}
therefore, n(E) = 18
And P(E) = p( getting two numbers whose sum is odd)
$$\eqalign{ & {\text{P}}\left( {\text{E}} \right) = \frac{{{\text{n}}\left( {\text{E}} \right)}}{{{\text{n}}\left( {\text{S}} \right)}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{18}}{{36}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{2} \cr} $$
Hence the answer is $$\frac{1}{2}$$

Given that, 45% play football; that is, P(F) = $$\frac{{45}}{{100}}$$ = $$\frac{{9}}{{20}}$$
30% play volleyball; that is, P(V) = $$\frac{{30}}{{100}}$$ = $$\frac{{6}}{{20}}$$
And, 15% play both volleyball and football; that is, P(F And V) = $$\frac{{15}}{{100}}$$ = $$\frac{{3}}{{20}}$$
Now, we have to find the probability that 1 student plays football or volleyball;
That we have to find, P(F or V)
We know that, P(F Or V) = P(F) + P(V) - P(F And V)
= $$\frac{{9}}{{20}}$$ + $$\frac{{6}}{{20}}$$ - $$\frac{{3}}{{20}}$$
= $$\frac{{12}}{{20}}$$
= $$\frac{{3}}{{5}}$$
Hence, the required probability $$\frac{{3}}{{5}}$$

Total 4 cases = [HH, TT, TH, HT]
Favourable cases = [HH, TH, HT]
Please note we need atmost one tail, not atleast one tail.
So probability = $$\frac{{3}}{{4}}$$

Total number of balls = (8 + 7 + 6) = 21
Let E = event that the ball drawn is neither blue nor green = event that the ball drawn is red.
Therefore, n(E) = 8
P(E) = $$\frac{{8}}{{21}}$$

Total cases = 10 + 20 = 30 Favourable cases = 20
So probability = $$\frac{{20}}{{30}}$$ = $$\frac{{2}}{{3}}$$

Total number of erasers in the box = 3 + 4 + 7 = 14
Let S be the sample space
Then, n(S) = number of ways of taking 5 out of 14
Therefore,
$$\eqalign{ & {\text{n}}\left( {\text{S}} \right) = {}^{14}{C_5} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{14 \times 13 \times 12 \times 11 \times 10}}{{2 \times 3 \times 4 \times 5}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = 14 \times 13 \times 11 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2002 \cr} $$
Let E be the event of getting all the 5 blue erasers
Therefore,
$$\eqalign{ & {\text{n}}\left( {\text{E}} \right) = {}^7{C_5} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{7 \times 6 \times 5 \times 4 \times 3}}{{2 \times 3 \times 4 \times 5}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 21 \cr} $$
Now, the required probability
$$\eqalign{ & \frac{{{\text{n}}\left( {\text{E}} \right)}}{{{\text{n}}\left( {\text{S}} \right)}} = \frac{{21}}{{2002}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{3}{{286}} \cr} $$

The numbers can be written in 5! different ways then n(S) = 5! = 120
Let E be the event that getting 4 digit number with 1 in the unit place and 2 in the tens place when the numbers 1, 2, 3, 4 and 5 are arranged at random.
Since each desired number is ending with 1, there is only 1 way
The Second place(tens) can now be filled with 2
Here also only 1 way to fill
The next place(hundreds) can now be filled by any of the remaining 3 numbers
So, there are 3 ways to filling that place
Then, the first place can now be filled by any of the remaining 2 numbers
So, there are 2 ways to fill
Therefore n(E) = 1 x 1 x 3 x 2 = 6
Now,
$$\eqalign{ & {\text{p}}\left( {\text{E}} \right) = \frac{{{\text{n}}\left( {\text{E}} \right)}}{{{\text{n}}\left( {\text{S}} \right)}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{6}{{120}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{{20}} \cr} $$
Hence the required probability is $$\frac{1}{{20}}$$

In a simultaneous throw of two dice, we have n(s) = 6 × 6 = 36
Let E = event of getting two numbers are same.
Then E = {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}
therefore, n(E) = 6
And p(E) = p(getting two numbers are same)
$$\eqalign{ & {\text{p}}\left( {\text{E}} \right) = \frac{{{\text{n}}\left( {\text{E}} \right)}}{{{\text{n}}\left( {\text{S}} \right)}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{6}{{36}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{6} \cr} $$
Hence the answer is $$\frac{{1}}{{6}}$$