Question

V Anand and Gary Kasparov play a series of 5 chess games. the probability that V Anand wins a game is 2/5 and the probability of Kasparov winning a game is 3/5. there is no probability of a draw. The series will be won by the person who wins 3 matches. Find the probability that Anand wins the series. (The series ends the moment when any of the two wins 3 matches.)

a) 992/3125 b) 273/625
c) 1021/3125 d) 1081/3125

Answer 1

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A OPTION IS THE ANSWER

Answer 2

Answer:

$\frac{992}{3125}$ is the probability that $\mathrm{V}$ Anand wins the series.

Step-by-step explanation:

Step 1:  The area of mathematics known as probability deals with numerical representations of the likelihood that an event will occur or that a statement is true. An event's probability is a number between 0 and 1, where, broadly speaking, 0 denotes the event's impossibility and 1 denotes certainty.

Step 2: According to the question,

As we know that -

V Anand and Gary Kasparov play a series of 5 chess games.

The probability that V Anand wins a game is $\frac{2}{5}$ and the probability of Kasparov winning a game is $\frac{3}{5}$.

First three wins by $\vee$ Anand $=\frac{8}{125}$

Step 3: 4 matches played V Anand win 3 out of 5 and in order to have all 5 matches Anand have to surely win 1 by earlier 4 matches he only won $2={ }^4 C_2 \times \frac{8}{125} \times \frac{9}{25}$

By adding all three $=\frac{8}{125}+\frac{72}{625}+\frac{432}{3125}$

$=\frac{992}{3125}$

Hence, $\frac{992}{3125}$ is the probability that $\mathrm{V}$ Anand wins the series.

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