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Probability

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Question : 91 of 127

Marks: +1, -0

If

7: 6

is in favour of

P

to survive 10 years more and 5: 3 in favour of Q to survive 10 years more, the probability that all least one of them will survive for 10 years more is :

$\frac{86}{104}$
$\frac{13}{26}$
$\frac{2}{26}$
$\frac{42}{52}$

Solution:

Concept:

The probability that at least one survives is

1 -

probability both die.

Explanation:

Given odds in favour of

P

surviving are

7:6

, so

P(\text{P survives}) = \frac{7}{7+6} = \frac{7}{13}

.

Thus

P(\text{P dies}) = 1 - \frac{7}{13} = \frac{6}{13}

.

Given odds in favour of

Q

surviving are

5:3

, so

P(\text{Q survives}) = \frac{5}{5+3} = \frac{5}{8}

.

Thus

P(\text{Q dies}) = 1 - \frac{5}{8} = \frac{3}{8}

.

Assuming independence,

P(\text{both die}) = \frac{6}{13} \times \frac{3}{8} = \frac{18}{104}

.

P(\text{at least one survives}) = 1 - \frac{18}{104} = \frac{86}{104}

.

Answer:

\frac{86}{104}

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