Question

An ice cream vendor sells three flavors: chocolate, strawberry, and vanilla. Forty five percent of the sales are chocolate, while 30% are strawberry, with the rest vanilla flavored. Sales are by the cone or the cup. The percentages of cones sales for chocolate, strawberry, and vanilla, are 75%, 60%, and 40%, respectively. For a randomly selected sale, define the following events:
a. chocolate chosen
b. strawberry chosen
c. vanilla chosen
d. ice cream on a cone
e. ice cream in a cup
Find the probability that the ice cream was sold on a cone and was strawberry flavor

Answer 1

Answer:

Step-by-step explanation:

Sales of the chocolate ice cream is 45%, 30% for strawberry and 25% (100%-45%-30%) of vanilla.

Percentages of cone sales for chocolate, strawberry and vanilla are 75%, 60% and 40%respectively.

A. Probability chocolate chosen: 45%=0.45

B. Probability strawberry chosen: 30%=0.30

C. Probability vanilla chosen:25%=0.25

D. Probability ice cream on a cone: 75%×45%+60%×30%+40%×25%

                                                          =0.75×0.45+0.60×0.30+0.40×0.25

                                                          =0.3375+0.18+0.1

                                                          =0.6175

E. Probability ice cream in a cup: 1- Probability ice cream in a cone

                                                     =1-0.6175

                                                     =0.3825

Probability that the ice cream sold on a cone and was strawberry flavoured is : 30%×60%

=0.30×0.60

=0.18

Answer 2

Answer: 0.18.

 

Step-by-step explanation: Let us define the following events

A= event that chocolate chosen,

B=event that strawberry chosen,

C=event that vanilla chosen,

D=event of choosing ice-cream on a cone

and

E=event of choosing ice-cream on a cup.

Then, according to the given information, we have

P(A)=0.45, P(B)=0.30, P(C)=0.25, P(A\D)=0.75, P(B\D)=0.60 and P(C\D)=0.40.

Therefore, the probability that the ice-cream was sold on a cone and was strawberry flavour is given by

$P(B\cap D)=P(B)\times 0.60\\\Rightarrow P(B\cap D)=0.30\times 0.60\\\Rightarrow P(B\cap D)=0.18.$

Thus, the required probability is 0.18.