In a consignment of 15 articles,it is found t | vedclass

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

Question

(C) Step $1$: Define the variables. Total articles $= 15$. Defective articles $= 3$. Non-defective articles $= 15 - 3 = 12$. Sample size $= 5$. We nee

Vedclass · Accepted Answer

$\frac{220}{1001}$

Vedclass · Answer

(C) Step $1$: Define the variables. Total articles $= 15$. Defective articles $= 3$. Non-defective articles $= 15 - 3 = 12$. Sample size $= 5$. We need the probability of choosing exactly $2$ defective articles in this sample. Step $2$: Use combinations to calculate possible outcomes. The probability is calculated using the hypergeometric distribution formula: $P(X = k) = \frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}}$ $1$. Number of ways to choose $2$ defective articles out of $3$: $\binom{3}{2} = \frac{3!}{2!1!} = 3$. $2$. Number of ways to choose $3$ non-defective articles out of $12$: $\binom{12}{3} = \frac{12 	imes 11 	imes 10}{3 	imes 2 	imes 1} = 220$. $3$. Total number of ways to choose $5$ articles out of $15$: $\binom{15}{5} = \frac{15 	imes 14 	imes 13 	imes 12 	imes 11}{5 	imes 4 	imes 3 	imes 2 	imes 1} = 3003$. Step $3$: Calculate the probability. $P = \frac{3 	imes 220}{3003} = \frac{660}{3003} = \frac{220}{1001}$. Thus,the correct answer is $\frac{220}{1001}$.

In a consignment of $15$ articles,it is found that $3$ are defective. If a sample of $5$ articles is chosen at random from it,then the probability of having $2$ defective articles is

Similar Questions

Two persons $A$ and $B$ have respectively $n + 1$ and $n$ coins which they toss simultaneously. Then the probability that $A$ will have more heads than $B$ is

Four cards are drawn one by one from a well-shuffled pack of $52$ cards. What is the probability that all four cards belong to the same suit?

From the first $100$ natural numbers, two numbers $a$ and $b$ are selected randomly without replacement. If the probability that $a-b \ge 10$ is $\frac{m}{n}$, with $\gcd(m, n)=1$, then $m+n$ is equal to:

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

In a shoe rack,there are $4$ pairs of shoes. If $4$ shoes are drawn one after the other at random without replacement,what is the probability of getting at least one correct pair of shoes among the four shoes drawn?

Vedclass Test Series

Exam Paper Generator

Online Exam Module