Question

a fair coin is flipped. if the flip results in a head, then a fruit is selected from a basket containing 8 apples, 2 bananas, and 6 cantaloups. if the flip results in a tail, then a fruit is selected from a basket containing 6 apples and 4 bananas. what is the probability that the flip resulted in tails, given that the fruit selexted is a banana g

Answer 1

Solution :

We have given two baskets :

$H_1$ : 8 apples + 2 bananas + 6 cantaloupes = 16 fruits

$H_2$ : 6 apples + 4 bananas = 10 fruits

A fair coin is made to flipped. If the $\text{flip}$ results is head, then the fruit is selected from a basket $H_1$.

If the flip results in tail, then the fruit is selected from the basket $H_2$.

Probability of head P(H) = $1/2$

Probability of tail P(T) = $1/2$

if given event is :

B = selected fruit is BANANA

We have to calculate : P(T|B)

Probability of banana if the flip results is head P(B|H) = $\frac{2}{16}$

Probability of banana if the flip results is tail P(B|T) = $\frac{4}{10}$

From the Bayes' theorem :

Probability of flip results is tail when selected fruit is BANANA.

$P(T|B) = \frac{P(B|T)\ P(T)}{P(B|T) \ P(T) + P(B|H)\ P(H)}$

            $=\frac{\frac{4}{10} \times \frac{1}{2}} {\frac{4}{10} \times \frac{1}{2} + \frac{2}{16} \times \frac{1}{2}}$

            $=\frac{\frac{1}{5}}{\frac{1}{5}+\frac{1}{16}}$

            $=\frac{\frac{1}{5}}{\frac{21}{80}}$

            $=\frac{16}{21}$

∴  $P(\ T|B\ )=\frac{16}{21}$