Find P(E | F) for a die thrown three times,wh | vedclass

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(A) When a die is thrown three times,the total number of outcomes in the sample space is $6 	imes 6 	imes 6 = 216$.Event $F$ is defined as $6$ and $

Vedclass · Accepted Answer

$\frac{1}{6}$

Vedclass · Answer

(A) When a die is thrown three times,the total number of outcomes in the sample space is $6 	imes 6 	imes 6 = 216$.Event $F$ is defined as $6$ and $5$ appearing on the first two tosses,respectively:$F = \{(6, 5, 1), (6, 5, 2), (6, 5, 3), (6, 5, 4), (6, 5, 5), (6, 5, 6)\}$.The number of elements in $F$ is $n(F) = 6$.Event $E$ is defined as $4$ appearing on the third toss.The intersection $E \cap F$ consists of outcomes where $6$ and $5$ appear on the first two tosses and $4$ appears on the third:$E \cap F = \{(6, 5, 4)\}$.The number of elements in $E \cap F$ is $n(E \cap F) = 1$.The conditional probability is given by $P(E | F) = \frac{n(E \cap F)}{n(F)}$.$P(E | F) = \frac{1}{6}$.

Find $P(E | F)$ for a die thrown three times,where $E: 4$ appears on the third toss,and $F: 6$ and $5$ appear respectively on the first two tosses.

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