Two persons A and B throw a (fair)die (six-faced cube with faces numbered from 1 to 6 ) alternately,

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14.Probability

normal

Two persons $A$ and $B$ throw a (fair)die (six-faced cube with faces numbered from $1$ to $6$ ) alternately, starting with $A$. The first person to get an outcome different from the previous one by the opponent wins. The probability that $B$ wins is

$\frac{5}{6}$

$\frac{6}{7}$

$\frac{7}{8}$

$\frac{8}{9}$

(KVPY-2014)

Solution

(b)

Given, Probability of win $=\frac{5}{6}$

Probability of loss $=\frac{1}{6}$

Probability of $B$ wins =

$P(B)+P(\overline{B A} B)+P(\bar{B} \bar{A} \bar{B} \bar{A} B)+\ldots$

$=\frac{5}{6}+\frac{1}{6} \times \frac{1}{6} \times \frac{5}{6}+\frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} \times \frac{5}{6}+\ldots$

$=\frac{5}{6}\left(1+\frac{1}{6^2}+\frac{1}{6^4}+\frac{1}{6^6}+\ldots\right)$

$=\frac{5}{6}\left(\frac{1}{1-\frac{1}{36}}\right)=\frac{30}{35}=\frac{6}{7}$

Standard 11

Mathematics

If $A$ and $B$ are two events of a random experiment, $P\,(A) = 0.25$, $P\,(B) = 0.5$ and $P\,(A \cap B) = 0.15,$ then $P\,(A \cap \bar B) = $

easy

The probability of solving a question by three students are $\frac{1}{2},\,\,\frac{1}{4},\,\,\frac{1}{6}$ respectively. Probability of question is being solved will be

medium

The probabilities of three events $A , B$ and $C$ are given by $P ( A )=0.6, P ( B )=0.4$ and $P ( C )=0.5$ If $P ( A \cup B )=0.8, P ( A \cap C )=0.3, P ( A \cap B \cap$ $C)=0.2, P(B \cap C)=\beta$ and $P(A \cup B \cup C)=\alpha$ where $0.85 \leq \alpha \leq 0.95,$ then $\beta$ lies in the interval

hard

(JEE MAIN-2020)

If $A$ and $B$ are two independent events, then $P\,(A + B) = $