Question

The odds against student X solving a business statistics problem are 8:6 and odds in favour of student Y solving the same problem are 14:16
a) What is the chance that the problem will be solved, if they try independently?
b) What is the probability that neither solves the problem?

Answer 1

We know that ,

Probability of happening of an event = Favourable outcome/total outcomes

The odds against student X solving a business statistics problem are 8:6

Probability that X doesn't solve the problem
$p(\bar X) = \frac{8}{14} = \frac{4}{7}$
Probability that X solve the problem
$p(X) = 1 - \frac{4}{7} \\ \\ p(X) = \frac{3}{7}$
odds in favour of student Y solving the same problem are 14:16

Probability that Y solve the problem

$p(Y) = \frac{14}{30} = \frac{7}{15}$
Probability that X doesn't solve the problem

$p(\bar Y) = \frac{8}{15}$

a) What is the chance that the problem will be solved, if they try independently?

1) X solve but Y does not solve

2) X does not solve but Y solved

3) X and Y both solve the problem

So, probability that the problem is solved
$= p(X) p(\bar Y)+p(Y) p(\bar X)+p(X)p(Y) \\\\= \frac{3}{7} \times \frac{8}{15} + \frac{4}{7} \times \frac{7}{15} + \frac{3}{7} \times \frac{7}{15} \\ \\ = \frac{24}{105} + \frac{28}{105} + \frac{21}{105} \\ \\ = \frac{24 + 28 + 21}{105} \\ \\ = \frac{73}{105}$
b) What is the probability that neither solves the problem?
$=p(\bar X) p(\bar Y)\\\\= \frac{4}{7} \times \frac{8}{15} \\ \\ = \frac{32}{105}$
Hope it helps you