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Probability

Section: Mathematics

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Question : 12 of 92

Marks: +1, -0

A die is thrown twice. Let A be the event of getting a prime number when the die is thrown first time and

B

be the event of getting an even number when the die is thrown second time. Then

P(A / \overline{B})

[AP EAPCET 23 May 2025 S2]

$\frac{1}{2}$
$\frac{2}{3}$
$\frac{1}{5}$
$\frac{3}{5}$

Solution:

Total outcomes

=\{1,2,3,4,5,6\}=6

Prime number

=\{2,3,5\}=3

So,

P(A)=\frac{3}{6}=\frac{1}{2}

and even numbers

=\{2,4,6\}=3

\therefore P(B)=\frac{3}{6}=\frac{1}{2}

Thus,

P(\overline{B})=1-P(B)=1-\frac{1}{2}=\frac{1}{2}

Total sample for 2 throws

=6 \times 6=36

Then,

P(A / \overline{B})=\frac{P(A \cap \overline{B})}{P(\overline{B})}

First throw is prime

=\{2,3,5\}

Second throw is not even

=\{1,3,5\}

Pairs where first is in

\{2,3,5\}

and second is in

\{1,3,5\}=3 \times 3=9

outcomes

So,

P(A \cap \overline{B})=\frac{9}{36}=\frac{1}{4}

Therefore,

P(A / \overline{B})=\frac{\frac{1}{4}}{\frac{1}{2}}=\frac{1}{2}

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