Question

Consider rolling two fair dice and observing the number of spots on the resulting upward face of each one. Letting A be the event that at least one of the dice results in 6 spots on its upward face, and E be the event that exactly one of the dice results in 6 spots on its upward face, give the value of P(E|A). (Note: This value can be easily obtained by looking at p. 2.3 of the class notes and using a reduced sample space. In fact, such an approach makes the solution so trivial, that for this problem, you don’t have to give any justification for your answer.)

Answer 1

Answer:

$P(E|A)= \frac{10}{11}$

Step-by-step explanation:

Given

Two rolls of die

$E \to$ one of the outcomes is 6

$A \to$ atleast one is 6

Required

P(E|A)

First, list out the outcome of each

$E = \{(1,6),(2,6),(3,6),(4,6),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5)\}$

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

So:

$P(E|A)= \frac{n(E\ n\ A)}{n(A)}$

Where:

$E\ n\ A = \{(1,6),(2,6),(3,6),(4,6),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5)\}$

$n(E\ n\ A) = 10$

$n(A) = 11$

So:

$P(E|A)= \frac{10}{11}$