Probability - Aptitude MCQ Questions and Solutions with Explanations

S = {(1, 5), (1, 6), (1, 7), (2, 5), (2, 6), (2, 7), (3, 5), (3, 6), (3, 7), (4, 5), (4, 6), (4, 7)}
Total element n(S) = 12
xy will be even when even x or y or both will be even.
Events of x, y being even is E.
E = {(1, 6), (2, 5), (2, 6), (2, 7), (3, 6), (4, 5), (4, 6),(4, 7)}
n(E) = 8
So, Probability
$$\eqalign{ & P = \frac{{n(E)}}{{n(S)}} \cr & P = \frac{8}{{12}} \cr & P = \frac{2}{3} \cr} $$

$$|X|$$ can take 7 values.
To get $$|X| < 2$$   (i.e., -2 < x < + 2) take X = {-1, 0, 1}
$$P\left( {|X| < 2} \right) = $$   $$\frac{{{\text{Favourable Cases}}}}{{{\text{Total Cases}}}}$$
$$ = \frac{3}{7}$$

Required probability = P(X) + P(Y) + P(Z) (all the events are mutually exclusive)
$$\eqalign{ & = \frac{1}{6} + \frac{1}{{10}} + \frac{1}{8} \cr & = \frac{{47}}{{120}} \cr} $$

$$P({\text{odd}}) = P({\text{even}}) = \frac{1}{2}$$     (because there are 50 odd and 50 even numbers)
Sum or the three numbers can be odd only under the following 4 scenarios:
$$\eqalign{ & {\text{odd}} + {\text{odd}} + {\text{odd}} \cr & = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \cr & = \frac{1}{8} \cr & {\text{odd}} + {\text{even}} + {\text{even}} \cr & = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \cr & = \frac{1}{8} \cr & {\text{even}} + {\text{odd}} + {\text{even}} \cr & = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \cr & = \frac{1}{8} \cr & {\text{even}} + {\text{even}} + {\text{odd}} \cr & = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \cr & = \frac{1}{8} \cr} $$
Other combinations of odd and even will give even numbers.
Adding up the 4 scenarios above:
$$\eqalign{ & = \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} \cr & = \frac{4}{8} \cr & = \frac{1}{2} \cr} $$

Total names in the lottery,
$$\eqalign{ & = 3 \times 100 + 2 \times 150 + 200 \cr & = 800 \cr} $$
Number of Year-III's names,
$$\eqalign{ & = 3 \times 100 \cr & = 300 \cr} $$
Required probability,
$$\eqalign{ & = \frac{{300}}{{800}} \cr & = \frac{3}{8} \cr} $$

Probability of all three balls being black
$$\eqalign{ & = \frac{{^5{C_3}}}{{^9{C_3}}} \cr & = \frac{5}{{42}} \cr} $$
Probability that three balls are not black
$$\eqalign{ & = 1 - \frac{5}{{42}} \cr & = \frac{{37}}{{42}} \cr} $$
Hence, odds against these ball being black
$$\eqalign{ & = \left( {\frac{{37}}{{42}}} \right):\left( {\frac{5}{{42}}} \right) \cr & = 37:5 \cr} $$

Let A be the event that ball selected from the first bag is red and ball selected from second bag is green.
Let B be the event that ball selected from the first bag is green and ball selected from second bag is red.
$$\eqalign{ & P\left( A \right) = \left( {\frac{5}{8}} \right) \times \left( {\frac{6}{{10}}} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{3}{8}\,{\text{and}} \cr & P\left( B \right) = \left( {\frac{3}{8}} \right) \times \left( {\frac{4}{{10}}} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{3}{{20}} \cr} $$
Hence, required probability,
$$\eqalign{ & = P(A) + P(B) \cr & = \frac{3}{8} + \frac{3}{{20}} \cr & = \frac{{21}}{{40}} \cr} $$

P(winning prize at least on one ticket)
= 1 - P("Losing on all tickets")
= 1 - (0.8)⁴ = (1 + (0.8)²)(1 - (0.8)²)
= (1.64)(0.36)
= 0.5904

Out of nine, five are good and four are defective.
Required probability
$$\eqalign{ & = \frac{{{}^4{C_4}}}{{{}^9{C_4}}} \cr & = \frac{1}{{126}} \cr} $$

The number of exhaustive outcomes is 36.
Let E be the event of getting an even number on one die and an odd number on the other.
Let the event of getting either both even or both odd
Then
$$\eqalign{ & P\left( {\overline E } \right) = \frac{{18}}{{36}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{2} \cr & P\left( E \right) = 1 - \frac{1}{2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{2} \cr} $$