If Mohan has 3 tickets of a lottery containin | vedclass

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

Question

(A) Total number of tickets $= 3 + 9 = 12$.Mohan selects $3$ tickets out of $12$. The total number of ways to select $3$ tickets is ${}^{12}C_3 = \fra

Vedclass · Accepted Answer

$\frac{34}{55}$

Vedclass · Answer

(A) Total number of tickets $= 3 + 9 = 12$.Mohan selects $3$ tickets out of $12$. The total number of ways to select $3$ tickets is ${}^{12}C_3 = \frac{12 	imes 11 	imes 10}{3 	imes 2 	imes 1} = 220$.The probability of winning at least one prize is $1 - P(	ext{winning no prize})$.Winning no prize means all $3$ tickets are selected from the $9$ blanks.The number of ways to select $3$ blanks out of $9$ is ${}^9C_3 = \frac{9 	imes 8 	imes 7}{3 	imes 2 	imes 1} = 84$.The probability of winning no prize is $\frac{84}{220} = \frac{21}{55}$.Therefore,the probability of winning at least one prize is $1 - \frac{21}{55} = \frac{34}{55}$.

If Mohan has $3$ tickets of a lottery containing $3$ prizes and $9$ blanks,then his chance of winning at least one prize is

Similar Questions

$A$ card is drawn from a well-shuffled pack of $52$ cards. The probability of getting a queen of clubs or a king of hearts is:

$A$ box contains $36$ tickets numbered $1$ to $36$. One ticket is drawn at random. Find the probability that the number on the ticket is either divisible by $3$ or is a perfect square.

In a single throw of $2$ dice,find the probability of getting a total of $11$.

Let $A, B,$ and $C$ be three mutually independent events. Consider the two statements $S_1$ and $S_2$:
$S_1: A$ and $B \cup C$ are independent.
$S_2: A$ and $B \cap C$ are independent.
Then:

$A$ and $B$ are two independent events such that $P(A) = 1/2$ and $P(B) = 1/3$. Then $P$ (neither $A$ nor $B$) is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module