Answer to Question #208976 in Statistics and Probability for mimi

Question

A takeaway pizza shop conducted analysis of the takeaway meals ordered, and found that 40%

of orders were for a meat pizza with a drink, 20% for a vegetarian pizza with a drink, 10% for a

meat pizza without a drink, and 30% for a vegetarian pizza without a drink. If a takeaway order is

randomly selected, find the probability that it is for:

(a) A meat pizza without a drink. (2)

(b) An order with a drink. (2)

(c) A meat pizza. (2)

(d) A meat pizza or an order with a drink. (3)

(e) A meat pizza, if the order included a drink. (2)

(f) An order with a drink, if a vegetarian pizza was ordered.

Answer 1

(a)

$$P(M\cap D')=0.1$$

(b)

$$P(D)=P(M\cap D)+P(V\cap D)$$

$$=0.4+0.2=0.6$$

(c)

$$P(M)=P(M\cap D)+P(M\cap D')$$

$$=0.4+0.1=0.5$$

(d)

$$P(M\cup D)=1-P(V\cap D')$$

$$=1-0.3=0.7$$

Or

$$P(M\cup D)=P(M)+P(V\cap D)$$

$$=0.5+0.2=0.7$$

(e)

$$P(M|D)=\dfrac{P(M\cap D)}{P(D)}=\dfrac{0.4}{0.6}=\dfrac{2}{3}$$

(f)

$$P(V)=P(V\cap D)+P(V\cap D')$$

$$=0.2+0.3=0.5$$

$$P(D|V)=\dfrac{P(V\cap D)}{P(V)}=\dfrac{0.2}{0.5}=0.4$$