For a biased die,the probabilities for differ | vedclass

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Question

(A) Let $E_1$ be the event that face $1$ turns up and $E_2$ be the event that face $2$ turns up.Given probabilities are $P(E_1) = 0.1$ and $P(E_2) = 0

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$\frac{5}{21}$

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(A) Let $E_1$ be the event that face $1$ turns up and $E_2$ be the event that face $2$ turns up.Given probabilities are $P(E_1) = 0.1$ and $P(E_2) = 0.32$.We are given that either face $1$ or $2$ has turned up. Let this event be $A = E_1 \cup E_2$.Since $E_1$ and $E_2$ are mutually exclusive,$P(A) = P(E_1) + P(E_2) = 0.1 + 0.32 = 0.42$.We need to find the conditional probability $P(E_1 | A)$,which is the probability that face $1$ has turned up given that either $1$ or $2$ has turned up.$P(E_1 | A) = \frac{P(E_1 \cap A)}{P(A)} = \frac{P(E_1)}{P(A)} = \frac{0.1}{0.42}$.Simplifying the fraction: $\frac{0.1}{0.42} = \frac{10}{42} = \frac{5}{21}$.

$Face$	$1$	$2$	$3$	$4$	$5$	$6$
$Probability$	$0.1$	$0.32$	$0.21$	$0.15$	$0.05$	$0.17$

For a biased die,the probabilities for different faces to turn up are given below:

$Face$ $1$ $2$ $3$ $4$ $5$ $6$

$Probability$ $0.1$ $0.32$ $0.21$ $0.15$ $0.05$ $0.17$

The die is tossed and you are told that either face $1$ or $2$ has turned up. Then the probability that it is face $1$ is:

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For a biased die,the probabilities for different faces to turn up are given below: $Face$ $1$ $2$ $3$ $4$ $5$ $6$ $Probability$ $0.1$ $0.32$ $0.21$ $0.15$ $0.05$ $0.17$ The die is tossed and you are told that either face $1$ or $2$ has turned up. Then the probability that it is face $1$ is: