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

Seven balls are drawn simultaneously from a bag containing $5$ white and $6$ green balls. The probability of drawing $3$ white and $4$ green balls is:

Find the probability that when a hand of $7$ cards is drawn from a well-shuffled deck of $52$ cards,it contains all $4$ Kings.

From a well-shuffled deck of $52$ cards,$13$ cards are selected at random. The probability that the selected $13$ cards contain exactly $4$ kings is ....

Three cards are drawn at random from a pack of $52$ cards. What is the probability of drawing three aces?

$3$ balls are drawn one after the other without replacement from an urn containing $4$ red, $5$ blue and $6$ yellow balls. The probability of getting three different coloured balls is

Vedclass Test Series

Exam Paper Generator

Online Exam Module