Two unbiased dice are thrown. The probability that the sum of the numbers appearing on the top face of two dice is greater than $7$ if $4$ appear on the top face of the first dice is...

\frac{1}{3}

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\frac{1}{2}

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\frac{1}{12}

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\frac{1}{6}

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Solution

The correct option is A $\frac{1}{3}$
$\begin{matrix} \begin{matrix} c o m b i n a t i o n s = (4, 1) (4, 2) (4, 3) (4, 4) \end{matrix}} \leq 7 (4, 5) (4, 6)} > 7 P (> 7) = \frac{1}{3} \end{matrix}$

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Two unbiased dice are thrown. The probability that the sum of the numbers appearing on the top face of two dice is greater than 7 if 4 appear on the top face of the first dice is...

The correct option is A 13combinations=(4,1)(4,2)(4,3)(4,4)}≤7(4,5)(4,6)}>7P(>7)=13

Two unbiased dice are thrown. The probability that the sum of the numbers appearing on the top face of two dice is greater than $7$ if $4$ appear on the top face of the first dice is...

The correct option is A $\frac{1}{3}$
$\begin{matrix} \begin{matrix} c o m b i n a t i o n s = (4, 1) (4, 2) (4, 3) (4, 4) \end{matrix}} \leq 7 (4, 5) (4, 6)} > 7 P (> 7) = \frac{1}{3} \end{matrix}$