Answer to Question #86649 in Statistics and Probability for Bob

Question

Suppose we have the following facts about customers at a bakery:

44% of customers buy bread
25% of customers buy a cupcake
58% of customers buy bread or a donut
11% of customers buy bread and a cupcake
8% of customers buy bread and a donut
10% of customers buy a cupcake and a donut
5% of customers buy all three items

What is the probability that the selected person buys a donut?

What is the probability that the selected person buys bread if we know they don't buy a cupcake?

What is the probability that the selected person buys bread if we know they buy a cupcake and a donut?

In a random sample of 5 customers, what is the probability that 2 or fewer of them buy bread?

Answer 1

Let B denote the event that the customer buys a bread, C denote the event that the customer buys a cupcake, and D denote the event that the customer buys a donut.

Then

$$P(B)=0.44, P(C)=0.25, P(B \cup D)=0.58,$$ $$P(B \cap C)=0.11, P(B \cap D)=0.08, P(C \cap D)=0.10, P(B \cap C \cap D)=0.05$$

Venn diagram

$$P(B \cap C \cap \overline{D})=P(B \cap C)-P(B \cap C \cap D)=0.11-0.05=0.06$$ $$P(B \cap D \cap \overline{C})=P(B \cap D)-P(B \cap C \cap D)=0.08-0.05=0.03$$ $$P(C \cap D \cap \overline{B})=P(C \cap D)-P(B \cap C \cap D)=0.10-0.05=0.05$$

$$P(B \cap \overline{C} \cap \overline{D})=P(B)-P(B \cap C \cap \overline{D})-P(B \cap D \cap \overline{C})-P(B \cap C \cap D)=$$ $$=0.25-0.06-0.05-0.05=0.09$$

$$P(C \cap \overline{B} \cap \overline{D})=P(C)-P(B \cap C \cap \overline{D})-P(C \cap D \cap \overline{B})-P(B \cap C \cap D)=$$ $$=0.25-0.06-0.05-0.05=0.09$$

$$P(D \cap \overline{B} \cap \overline{C})=P(B \cup D)-P(B \cap \overline{C} \cap \overline{D})-P(B \cap D \cap \overline{C})-P(B \cap C \cap D)-$$ $$-P(C \cap D \cap \overline{B})-P(B \cap C \cap \overline{D})=$$ $$=0.58-0.30-0.03-0.05-0.05-0.06=0.09$$

The probability that the selected person buys a donut is

$$P(D)=P(B \cup D)-P(B \cap \overline{C} \cap \overline{D})-P(B \cap C \cap \overline{D})=0.58-0.30-0.06=0.22$$

The probability that the selected person buys bread if we know they don't buy a cupcake is

$$P(B| \overline{C})=P(B \cap \overline{C})/P(\overline{C})=$$ $$=(P(B)-P(B \cap C))/(1-P(C))=(0.44-0.11)/(1-0.25)=0.44$$

The probability that the selected person buys bread if we know they buy a cupcake and a donut is

$$P(B|(C \cap D))=P(B \cap C \cap D)/P(C \cap D)=0.05/0.10=0.5$$

In a random sample of 5 customers, what is the probability that 2 or fewer of them buy bread? 

B has a binomial distribution

$$Bin (n=5, p=0.44)$$

$$P(B\le 2)=5!/(0!(5-0)!)*(0.44)^0*(1-0.44)^{5-0}+5!/(1!(5-1)!)*(0.44)^1*(1-0.44)^{5-1}+$$ $$+5!/(2!(5-2)!)*(0.44)^2*(1-0.44)^{5-2}=0.6114$$