Probability that the product of the out | vedclass

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Question

Let us divide the outcomes into three separate cases :

$I.$ Exactly one has even outcome

${P_1} = \frac{{{\,^3}{C_1} 	imes 1 	imes 3 	imes 3}

Vedclass · Accepted Answer

$\frac{5}{8}$

Vedclass · Answer

Let us divide the outcomes into three separate cases :

$I.$ Exactly one has even outcome

${P_1} = \frac{{{\,^3}{C_1} 	imes 1 	imes 3 	imes 3}}{{6 	imes 6 	imes 6}} = \frac{1}{8}$

$II.$ Exactly two have even outcomes

$P_{1}=\frac{^{3} C_{1} 	imes 3 	imes 3 	imes 3}{6 	imes 6 	imes 6}=\frac{3}{8}$

$III.$ Exactly have even outcomes

$P_{3}=\frac{3 	imes 3 	imes 3}{6 	imes 6 	imes 6}=\frac{1}{8}$

$	herefore \mathrm{P}=\mathrm{P}_{1}+\mathrm{P}_{2}+\mathrm{P}_{3}=\frac{5}{8}$

Probability that the product of the outcomes when three dice are rolled simultaneously is divisible by $4$ is equal to

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