A dice is thrown twice. probability of getting 4, 5 or 6 in first throw and 1, 2, 3 or 4 in second t

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14.Probability

easy

A dice is thrown twice. The probability of getting $4, 5$ or $6$ in the first throw and $1, 2, 3$ or $4$ in the second throw is

A

$1$

B

$\frac{1}{3}$

C

$\frac{7}{{36}}$

D

None of these

Solution

(b) Let $P(A)$ and $P(B)$ be the probability of the events then $P(A\,\,{\rm{and}}\,\,B)$

$= P(A)\,.\,P(B) = \frac{1}{2} \times \frac{2}{3} = \frac{1}{3}.$

Standard 11

Mathematics

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