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Odds in Favor and Odds against an Event

Two different...

Question

Two different digits are chosen at random fromthe set $1, 2, 3, 4, 5, 6, 7, 8.$ Then the probability that both the digits exceed $5$ is

A

\frac{3}{28}

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B

\frac{5}{28}

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C

\frac{9}{28}

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D

\frac{7}{28}

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Solution

The correct option is A $\frac{3}{28}$
Possible numbers are $6, 7, 8$ .
$\Rightarrow P =^{3} C_{2} /^{8} C_{2} = 3 / 28$

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