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adelina 88 [10]
3 years ago
8

Consider rolling two fair dice and observing the number of spots on the resulting upward face of each one. Letting A be the even

t 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.)
Mathematics
1 answer:
Allushta [10]3 years ago
7 0

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}

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