Question

3) When I lived in California I had a small lemon tree in the front yard. If we had rain in the summer (rare=P=.2) it would yield up to 10 lemons, distributed with a binomial distribution, N= 10, P=0.6. If it is perfectly dry (most of the time P=0.8) it the distribution would be binomial with N=6, P=0.4 a) If you get 3 lemons what the probability that it rained. b) If you have 4 lemons, what is the probability that it rained.

Answer 1

a) Let X be the number of lemons

We have to find

P(rain I X =3 )

We know that

P(rain I X =3) = P( X =3 I rain ) .P(rain) / P(X=3)

P(dry) =0.8

P(rain) =0.2

When it rains , X follow Binomial with n=10, p=0.6

= 0.0425

When its dry , X follow Binomial with n= 6 , p=0.4

=0.2765

Thus , P(X=3) = P( X =3 I rain ) .P(rain) + P(X=3 I dry ) .P(dry)

= 0.0425* 0.2 + 0.2765* 0.8

= 0.2297

P(rain I X =3) = 0.0425*0.2 / 0.2297

= 0.0370

Probability that it rained given that we get 3 lemons = 0.0370

b)

We have to find

P(rain I X =4 )

We know that

P(rain I X =4) = P( X =4 I rain ) .P(rain) / P(X=4)

When it rains , X follow Binomial with n=10, p=0.6

= 0.1115

When its dry , X follow Binomial with n= 6 , p=0.4

=0.1382

Thus , P(X=4) = P( X =4 I rain ) .P(rain) + P(X=4 I dry ) .P(dry)

= 0.1329

Therefore ,

P(rain I X =4) = 0.1115*0.2 / 0.1329

= 0.1678  

Probability that it rained given that we get 4 lemons = 0.1678