The numbers 2, 3, 5, 7, 11, 13 are written on | vedclass

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Question

(A) The given numbers are $S = \{2, 3, 5, 7, 11, 13\}$.Total number of ways to choose $3$ chits out of $6$ is given by $^6C_3 = \frac{6 	imes 5 	ime

Vedclass · Accepted Answer

$\frac{7}{20}$

Vedclass · Answer

(A) The given numbers are $S = \{2, 3, 5, 7, 11, 13\}$.Total number of ways to choose $3$ chits out of $6$ is given by $^6C_3 = \frac{6 	imes 5 	imes 4}{3 	imes 2 	imes 1} = 20$.We classify the numbers based on their remainder when divided by $3$:- Remainder $0$: ${3}$ (count $n_0 = 1$)- Remainder $1$: ${7, 13}$ (count $n_1 = 2$)- Remainder $2$: ${2, 5, 11}$ (count $n_2 = 3$)For the sum of $3$ numbers to be divisible by $3$,the possible combinations of remainders $(r_1, r_2, r_3)$ are:$1$. $(0, 1, 2)$: Choose one from each set. Number of ways $= 1 	imes 2 	imes 3 = 6$.$2$. $(0, 0, 0)$: Not possible as we only have one number with remainder $0$.$3$. $(1, 1, 1)$: Not possible as we only have two numbers with remainder $1$.$4$. $(2, 2, 2)$: Choose three from the set with remainder $2$. Number of ways $= ^3C_3 = 1$.Total favorable outcomes $= 6 + 1 = 7$.Therefore,the required probability $= \frac{7}{20}$.

The numbers $2, 3, 5, 7, 11, 13$ are written on six distinct paper chits. If $3$ of them are chosen at random,then the probability that the sum of the numbers on the obtained chits is divisible by $3$ is:

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