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KEAM 2019 Math Paper

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Question : 53 of 121

Marks: +1, -0

Ram and Shyam throw a dice. The probability that Ram's throw is not greater than shyam's is

$\; \frac{1}{3}$
$\; \frac{1}{6}$
$\; \frac{7}{12}$
$\; \frac{2}{3}$
None of these

Solution:

According to the question, If

B

throws 1, then

A

is allowed to throw only 1 , if

B

throws 2, then

A

can throw 1 and 2 and soon. Hence, required probability.

\; \frac{1}{6} \cdot \; \frac{1}{6} + \; \frac{1}{6} \cdot \; \frac{2}{6} + \; \frac{1}{6} \cdot \; \frac{3}{6} + \; \frac{1}{6} \cdot \; \frac{4}{6} \cdot \; \frac{1}{6} \cdot \; \frac{5}{6} + \; \frac{1}{6} \cdot \; \frac{6}{6}

= \; \frac{1}{6 \times 6} (1+2+3+4+5+6)

= \; \frac{6 \times 7}{6 \times 6 \times 2} = \; \frac{7}{12}

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