A box contains 30 toys of the same size,in wh | vedclass

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

Question

(B) Total toys = $30$. White toys = $10$. Blue toys = $30 - 10 = 20$. Probability of drawing a white toy $(p)$ = $\frac{10}{30} = \frac{1}{3}$. Probab

Vedclass · Accepted Answer

$\left(\frac{8}{9}\right)^2$

Vedclass · Answer

(B) Total toys = $30$. White toys = $10$. Blue toys = $30 - 10 = 20$. Probability of drawing a white toy $(p)$ = $\frac{10}{30} = \frac{1}{3}$. Probability of drawing a blue toy $(q)$ = $1 - p = 1 - \frac{1}{3} = \frac{2}{3}$. Since the toys are replaced,this follows a binomial distribution $B(n, p)$ with $n = 5$ and $p = \frac{1}{3}$. We need the probability of getting at most $2$ white toys,which is $P(X \le 2) = P(X=0) + P(X=1) + P(X=2)$. $P(X=k) = {}^nC_k p^k q^{n-k}$. $P(X=0) = {}^5C_0 (\frac{1}{3})^0 (\frac{2}{3})^5 = (\frac{2}{3})^5$. $P(X=1) = {}^5C_1 (\frac{1}{3})^1 (\frac{2}{3})^4 = 5 	imes \frac{1}{3} 	imes \frac{16}{81} = \frac{80}{243}$. $P(X=2) = {}^5C_2 (\frac{1}{3})^2 (\frac{2}{3})^3 = 10 	imes \frac{1}{9} 	imes \frac{8}{27} = \frac{80}{243}$. $P(X \le 2) = \frac{32}{243} + \frac{80}{243} + \frac{80}{243} = \frac{192}{243}$. Dividing numerator and denominator by $3$,we get $\frac{64}{81} = (\frac{8}{9})^2$.

$A$ box contains $30$ toys of the same size,in which $10$ toys are white and the remaining toys are blue. $A$ toy is drawn at random from the box and it is replaced in the box after noting down its colour. If $5$ toys are drawn in this way,then the probability of getting at most $2$ white toys is

Similar Questions

The probability that a person will develop immunity after vaccination is $0.8$. If $8$ people are given the vaccine,then the probability that all develop immunity is equal to:

It is known that $10 \%$ of certain articles manufactured are defective. What is the probability that in a random sample of $12$ such articles,$9$ are defective?

$A$ coin is biased so that a head is twice as likely to occur as a tail. If the coin is tossed $3$ times,then the probability of getting two tails and one head is-

$A$ man takes a step forward with probability $0.4$ and one step backward with probability $0.6$. Then the probability that at the end of eleven steps he is one step away from the starting point is

For a binomial distribution $B(n, p)$,if the mean $= 200$ and the standard deviation $= 10$,then the value of $n^2 + \frac{1}{p^2} + \frac{1}{q^2}$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module