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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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:

Similar Questions

Out of $40$ consecutive natural numbers,two are chosen at random. The probability that the sum of the numbers is odd is:

$A$ bag contains $6$ white and $4$ black balls. $A$ die is rolled once and the number of balls equal to the number obtained on the die are drawn from the bag at random. The probability that all the balls drawn are white is:

Seven chits are numbered $1$ to $7$. Three are drawn one by one with replacement. The probability that the least number on any selected chit is $5$ is:

$A$ and $B$ play $12$ games. $A$ wins $6$ times,$B$ wins $4$ times,and $2$ games are drawn. If $A$ and $B$ play a series of $3$ games,what is the probability that they win alternately?

If a coin is tossed seven times, then the probability of getting exactly three heads such that no two heads occur consecutively is

Vedclass Test Series

Exam Paper Generator

Online Exam Module