A die is thrown twice and the sum of the numb | vedclass

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(A) Let $F$ be the event that the sum of the numbers appearing is $6$. The sample space for $F$ is $F = \{(1,5), (2,4), (3,3), (4,2), (5,1)\}$. Thus,$

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$\frac{2}{5}$

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(A) Let $F$ be the event that the sum of the numbers appearing is $6$. The sample space for $F$ is $F = \{(1,5), (2,4), (3,3), (4,2), (5,1)\}$. Thus,$n(F) = 5$.Let $E$ be the event that the number $4$ appears at least once. We are interested in the event $E \cap F$,which represents the outcomes where the sum is $6$ $AND$ the number $4$ appears at least once.From the set $F$,the outcomes that contain at least one $4$ are $(2,4)$ and $(4,2)$.Thus,$E \cap F = \{(2,4), (4,2)\}$.Therefore,$n(E \cap F) = 2$.The conditional probability $P(E|F)$ is given by the formula:$P(E|F) = \frac{n(E \cap F)}{n(F)} = \frac{2}{5}$.

$Face$	$1$	$2$	$3$	$4$	$5$	$6$
$P(F)$	$0.2$	$0.22$	$0.11$	$0.25$	$0.05$	$0.17$

$A$ die is thrown twice and the sum of the numbers appearing is observed to be $6$. What is the conditional probability that the number $4$ has appeared at least once?

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