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Exercise 2.9

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(a)

P (A_{1} | B) = \frac{P (A_{1}) P (B | A_{1})}{P (B)} = \frac{P (A_{1})}{P (B)} = \frac{P (A_{2})}{P (B)} \frac{P (A_{2}) P (B | A_{2})}{P (B)} = P (A_{2} | B)

(b)

B

is implied by both

A_{1}

and

A_{2}

, knowing that

B

occured does not tip the probability of occurence in favor of either

A_{1}

A_{2}

For example, let $A_{1}$ be the event that the card in my hand is the Ace of Spades. Let $A_{2}$ be the event that the card in my hand is the Ace of Hearts. Let $B$ be the event that there are $3$ aces left in the deck.

$B$ is implied by both $A_{1}$ and $A_{2}$ , and $P (A_{1}) = P (A_{2})$ . Knowing that $B$ occured does not give one any information on whether they are holding the Ace of Spades or the Ace of Hearts, since $B$ would have occured in both cases. Thus, $P (A_{1} | B) = P (A_{2} | B)$ .

fifthist

2021-12-05 00:00

Exercise 2.9

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