A box contains 15 tickets numbered 1, 2, dots | vedclass

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

Question

(D) Let $X$ be the random variable representing the number on a drawn ticket. Since the tickets are drawn with replacement,each draw is independent.Fo

Vedclass · Accepted Answer

None of these

Vedclass · Answer

(D) Let $X$ be the random variable representing the number on a drawn ticket. Since the tickets are drawn with replacement,each draw is independent.For each draw,the probability that the number on the ticket is less than or equal to $9$ is $P(X \le 9) = \frac{9}{15} = \frac{3}{5}$.The probability that the greatest number on the $7$ drawn tickets is $9$ is given by $P(	ext{max} = 9) = P(	ext{max} \le 9) - P(	ext{max} \le 8)$.$P(	ext{max} \le 9) = (\frac{9}{15})^7 = (\frac{3}{5})^7$.$P(	ext{max} \le 8) = (\frac{8}{15})^7$.Therefore,the required probability is $(\frac{9}{15})^7 - (\frac{8}{15})^7 = (\frac{3}{5})^7 - (\frac{8}{15})^7$.Since this result is not among the given options,the correct choice is $D$.

$A$ box contains $15$ tickets numbered $1, 2, \dots, 15$. Seven tickets are drawn at random one after the other with replacement. The probability that the greatest number on a drawn ticket is $9$ is:

Similar Questions

Let $0 < P(A) < 1$,$0 < P(B) < 1$ and $P(A \cup B) = P(A) + P(B) - P(A)P(B).$ Then

If two smallest squares are chosen at random on a chessboard, then the probability of choosing these squares such that they do not have a side in common is:

$A$ bag contains $n + 1$ coins. Among these coins,one coin has heads on both sides,while all other coins are fair. $A$ coin is chosen at random from the bag. If the probability of getting a head when the chosen coin is tossed is $7/12$,what is the value of $n$?

If $E_1, E_2, \ldots, E_n$ are independent events such that $P(E_r) = \frac{1}{1+r}$ for $r = 1, 2, \ldots, n$, then the probability that at least one of $E_1, E_2, \ldots, E_n$ happens is

If $A$ and $B$ are independent events of a random experiment such that $P(A \cap B)=\frac{1}{6}$ and $P(\bar{A} \cap \bar{B})=\frac{1}{3}$, then $P(A)$ is equal to (Here, $\bar{E}$ is the complement of the event $E$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module