An unbiased die is tossed until a number grea | vedclass

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

Question

(B) Let $p$ be the probability of getting a number greater than $4$ (i.e.,$5$ or $6$) in a single toss.$p = \frac{2}{6} = \frac{1}{3}$.Let $q$ be the

Vedclass · Accepted Answer

$\frac{2}{5}$

Vedclass · Answer

(B) Let $p$ be the probability of getting a number greater than $4$ (i.e.,$5$ or $6$) in a single toss.$p = \frac{2}{6} = \frac{1}{3}$.Let $q$ be the probability of failure (i.e.,getting $1, 2, 3,$ or $4$).$q = 1 - p = 1 - \frac{1}{3} = \frac{2}{3}$.We need the probability that the success occurs on an even number of tosses (i.e.,$2^{nd}, 4^{th}, 6^{th}, \dots$ toss).The probability is given by the sum of the infinite geometric series:$P = P(FS) + P(FFFS) + P(FFFFFS) + \dots$$P = qp + q^3p + q^5p + \dots$This is a geometric progression with first term $a = qp$ and common ratio $r = q^2$.The sum of an infinite geometric series is $S = \frac{a}{1-r}$.$P = \frac{qp}{1 - q^2} = \frac{(\frac{2}{3})(\frac{1}{3})}{1 - (\frac{2}{3})^2} = \frac{\frac{2}{9}}{1 - \frac{4}{9}} = \frac{\frac{2}{9}}{\frac{5}{9}} = \frac{2}{5}$.

An unbiased die is tossed until a number greater than $4$ appears. The probability that an even number of tosses is needed is

Similar Questions

$3$ coins are tossed. Find the probability of getting at least $1$ head and $1$ tail.

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

Two dice are thrown simultaneously. What is the probability of obtaining a sum of the numbers less than $11$?

The probability that in a year of the $22^{nd}$ century chosen at random there will be $53$ Sundays is

$2$ dice are thrown. Find the odds against getting the sum $6$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module