Six faces of an unbiased die are numbered wit | vedclass

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(A) The faces of the die are $\{2, 3, 5, 7, 11, 13\}$.There is $1$ even number $(2)$ and $5$ odd numbers $(3, 5, 7, 11, 13)$.The sum of two numbers is

Vedclass · Accepted Answer

$\frac{5}{18}$

Vedclass · Answer

(A) The faces of the die are $\{2, 3, 5, 7, 11, 13\}$.There is $1$ even number $(2)$ and $5$ odd numbers $(3, 5, 7, 11, 13)$.The sum of two numbers is odd if one number is even and the other is odd.Let $E$ be the event of getting an even number and $O$ be the event of getting an odd number.$P(E) = \frac{1}{6}$ and $P(O) = \frac{5}{6}$.The sum is odd in two cases: (First die is even, Second die is odd) or (First die is odd, Second die is even).Required Probability $= P(E) 	imes P(O) + P(O) 	imes P(E)$$= \left(\frac{1}{6} 	imes \frac{5}{6}ight) + \left(\frac{5}{6} 	imes \frac{1}{6}ight)$$= \frac{5}{36} + \frac{5}{36} = \frac{10}{36} = \frac{5}{18}$.

Six faces of an unbiased die are numbered with $2, 3, 5, 7, 11$ and $13$. If two such dice are thrown, then the probability that the sum on the uppermost faces of the dice is an odd number is

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