Question

Two dice are thrown. Find the probability that the number appeared have a sum 8 if it is known that second dice always exhibits 4.

Answer 1

Please see the attachment

Answer 2

Answer:

$P(B/A)=\frac{1}{6}$          

Step-by-step explanation:

Given : Two dice are thrown.

To find : The probability that the number appeared have a sum 8 if it is known that second dice always exhibits 4 ?

Solution :

Consider the given event,

A = 4 appears on a second die

B = The sum of the number on two dice is 8.

i.e. $A=\{(1,4),(2,4),(3,4),(4,4),(5,4),(6,4)\}$

$B=\{(4,4),(3,5),(5,3),(2,6),(6,2)\}$

Now, $A\cap B=\{(4,4)\}$

The required probability that the number appeared have a sum 8 if it is known that second dice always exhibits 4 is given by,

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

$P(B/A)=\frac{1}{6}$