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Standard XII

Mathematics

Mutually Exclusive Events

A and B throw...

Question

A and B throw a pair of dice. A wins if he throws $6$ before B throws $7$ and B wins if he throws $7$ before A throws $6$ . If A begins, his chance of winning is?

\frac{5}{16}

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\frac{30}{61}

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\frac{35}{61}

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\frac{60}{61}

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Solution

The correct option is B $\frac{30}{61}$
The probability of A throwing ‘6’ is 5/36. So, the probability of A not throwing ‘6’ is 1 – 5/36 = 31/36. The probability of B throwing ‘7’ is 6/36. So, the probability of B not throwing ‘7’ is 1 – 6/36 = 30/36.

The probability that somebody wins is [1-{(31/36)(30/36)}]. So, The probability of winning of A provided that somebody wins
= (5/36)/[1-{(31/36)(30/36)}] = 30/61.

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A and B throw a pair of dice. A wins if he throws 6 before B throws 7 and B wins if he throws 7 before A throws 6. If A begins, his chance of winning is?

A and B throw a pair of dice. A wins if he throws $6$ before B throws $7$ and B wins if he throws $7$ before A throws $6$ . If A begins, his chance of winning is?