Probability

[Emily Ma]

The probability that Greta’s mother takes her shopping is 2/5.  When Greta goes shopping with her mother she gets an ice cream 70% of the time. When Greta does not go shopping with her mother she gets an ice cream 30% of the time.

 

Determine the probability that:

 

a Greta’s mother buys her an icecream when shopping.

 

b Greta went shopping with her mother, given that her mother buys her an icecream.

 

 

I keep on getting the wrong answer

[IB_taking_over]

Easiest way to get the answer is to create a tree diagram. 

                  p=.7  I

   p=.4  S < 

                  p=.3  I'

<

                  p=.3  I

  p=.6   S' <

                  p=.7  I'

 

a) S and I using the tree diagram we see the p of S=.4 and the p of I=.7 

.4*.7=.28 

28% of the time she sill go shopping and eat ice cream.

 

b) S given I. (For this you are going to want the IB formula packet.)

 

P(S&I)=p(I)*p(S given I)

 

.28/.7=p(s given I) = .4

 

part a should be correct and I think that's how you do part b

Edited April 1, 2015 by IB_taking_over

[L'estrange98]

Hi! For Part b you should use the formula for conditional probability P(A l B)= (A&B)/P(B).

I recommend to obtain first the probability  that Greeta has ob getting an ice cream. P(I)=(0.7*0.4)+(0.6*0.3)= (23/50)

Then you calculate the probability that Greeta went shopping and obtained an ice cream P(I&S)= (0.4*0.7)= 0.28

Thus,  P(S l I)=  P(S&I)/ P(I)  That is to say 0.28/(23/50) = 14/23