Respuesta :
Solution :
We have given two baskets :
[tex]$H_1$[/tex] : 8 apples + 2 bananas + 6 cantaloupes = 16 fruits
[tex]$H_2$[/tex] : 6 apples + 4 bananas = 10 fruits
A fair coin is made to flipped. If the [tex]\text{flip}[/tex] results is head, then the fruit is selected from a basket [tex]$H_1$[/tex].
If the flip results in tail, then the fruit is selected from the basket [tex]$H_2$[/tex].
Probability of head P(H) = [tex]1/2[/tex]
Probability of tail P(T) = [tex]1/2[/tex]
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) = [tex]$\frac{2}{16}$[/tex]
Probability of banana if the flip results is tail P(B|T) = [tex]$\frac{4}{10}$[/tex]
From the Bayes' theorem :
Probability of flip results is tail when selected fruit is BANANA.
[tex]$P(T|B) = \frac{P(B|T)\ P(T)}{P(B|T) \ P(T) + P(B|H)\ P(H)}$[/tex]
[tex]$=\frac{\frac{4}{10} \times \frac{1}{2}} {\frac{4}{10} \times \frac{1}{2} + \frac{2}{16} \times \frac{1}{2}}$[/tex]
[tex]$=\frac{\frac{1}{5}}{\frac{1}{5}+\frac{1}{16}}$[/tex]
[tex]$=\frac{\frac{1}{5}}{\frac{21}{80}}$[/tex]
[tex]$=\frac{16}{21}$[/tex]
∴ [tex]$P(\ T|B\ )=\frac{16}{21}$[/tex]
