Answer to Question #274077 in Statistics and Probability for Grace

Question

In a restaurant, customers may order any combination of chips, salad or onion rings. The probability that a customer chooses onion rings is 0.3, salad 0.4 chips and salad 0.15, chips and onion rings 0.55, all three 0.05, none 0.2.

calculate the probability a customer chooses

chips

exactly one item

Answer 1

Customers at a restaurant may order any combination of chips, salad or onion rings.

The probability that a customers chooses onion rings is 0.3, salad 0.4, chips and salad 0.15, chips and onion rings 0.15, salad or onion rings 0.55, all three 0.05, none 0.2.

Calculate the probability a customer chooses:

(i) chips (ii) chips only

Given

$$P(R)=0.3, P(S)=0.4, P(C\cap S)=0.15,$$

$$P(C\cap R)=0.15,P( S\cup R)=0.55,$$

$$P(C\cap S\cap R)=0.05,P((C\cup S\cup R)')=0.2$$

$$P(C\cup S \cup R)=1-P((C\cup S\cup R)')$$

$$=1-0.2=0.8$$

i)

$$P(C\cup S \cup R)=P(C)+P(S)+P(R)$$

$$-P(C\cap S)-P(C\cap R)-P(S\cap R)$$

$$+P(C\cap S\cap R)$$

$$P( S\cup R)=P(S)+P(R)-P( S\cap R)$$

$$0.55=0.4+0.3-P( S\cap R)$$

$$P( S\cap R)=0.15$$

$$0.8=P(C)+0.4+0.3$$

$$-0.15-0.15-0.15+0.05$$

$$P(C)=0.5$$

(ii)

$$P(exactly\ 1)=0.25+0.15+0.05=0.45$$