A die is loaded in such a way that each odd n | vedclass

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(A) Let the probability of each even number $(2, 4, 6)$ be $p$.Then the probability of each odd number $(1, 3, 5)$ is $2p$.Since the sum of all probab

Vedclass · Accepted Answer

$\frac{4}{9}$

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(A) Let the probability of each even number $(2, 4, 6)$ be $p$.Then the probability of each odd number $(1, 3, 5)$ is $2p$.Since the sum of all probabilities must be $1$,we have:$3(2p) + 3(p) = 1$$6p + 3p = 1$$9p = 1 \Rightarrow p = \frac{1}{9}$.The event $E$ is that a number greater than or equal to $4$ occurs,so $E = \{4, 5, 6\}$.$P(E) = P(4) + P(5) + P(6)$.Since $4$ and $6$ are even,$P(4) = p = \frac{1}{9}$ and $P(6) = p = \frac{1}{9}$.Since $5$ is odd,$P(5) = 2p = \frac{2}{9}$.$P(E) = \frac{1}{9} + \frac{2}{9} + \frac{1}{9} = \frac{4}{9}$.

$X=x$	$-3$	$0$	$1$	$\alpha$
$P(X=x)$	$\frac{1}{4}$	$K$	$\frac{1}{4}$	$\frac{1}{3}$

$A$ die is loaded in such a way that each odd number is twice as likely to occur as each even number. If $E$ is the event that a number greater than or equal to $4$ occurs on a single toss of the die,then $P(E)$ is equal to:

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