Six faces of an unbiased die are numbered wit | vedclass

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Question

(A) The numbers on 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

Vedclass · Accepted Answer

$\frac{5}{18}$

Vedclass · Answer

(A) The numbers on 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 and only if one number is even and the other is odd.Let $E$ be the event that the first die shows an even number and the second shows an odd number,or the first die shows an odd number and the second shows an even number.Probability of getting an even number $P(Even) = \frac{1}{6}$.Probability of getting an odd number $P(Odd) = \frac{5}{6}$.Required probability $= P(Even) 	imes P(Odd) + P(Odd) 	imes P(Even)$$= \left(\frac{1}{6} 	imes \frac{5}{6}ight) + \left(\frac{5}{6} 	imes \frac{1}{6}ight)$$= \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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