Question

Normal Calibri 1 (1) In a plzza restaurant, 80% of the customer ordered plzza. 70% of the customer ordered salad. If 60% of the customer ordered salad and ordered pizza, what is the probability that a randomly selected a customer will ord pizza or salad? [2] In a pizza restaurant, 80% of the customer ordered pizza. 70% of the customer ordered salad or soup. If 60% of the customer ordered pizza after they ordered salad o soup, what is the probability that a randomly selected a customer will order pizza and salad or soup? (3) a box contains 3 Red. 4 Green, and 2 Blue balls. (a) you draw 2 balls from the box, but do not replace the first ball in the box before drawing the second. Find the probability that the first ball is a Red and second ball is a Green. (b) Repeat part (a), but do replace the first ball in the box before the second.

Answer 1

Q1) We are given here that:
P( pizza ) = 0.8,
P( salad) = 0.7,
P( salad and pizza ) = 0.6

Using law of additivity of probability, we have here:
P( salad or pizza) = P(salad) + P(pizza) - P( salad and pizza ) = 0.8 + 0.7 - 0.6 = 0.9

Therefore 0.9 is the required probability here.

Q2) We are given here that:
P( pizza ) = 0.8,
P( salad or soup) = 0.7,
Now we are given here that 60% of the customers ordered pizza after they ordered salad or soup, therefore, we are given the conditional probability here as:
P( pizza | salad or soup) = 0.6

Using Bayes theorem here, we get here:
P( pizza and salad or soup) = P( pizza | salad or soup)P( salad or soup) = 0.6*0.7 = 0.42

Therefore 0.42 is the required probability here.

Q3) a) Probability that the first ball is red and second is green is computed here as:
= P(red)P(green given that first ball was red)

= (3/9)*(4/8)

= 1/6

= 0.1667

Therefore 0.1667 is the required probability here.

b) As we are drawing with replacement here, the probability here is computed as:
= P(red)P(green)

= (3/9)*(4/9)

= 0.1481

Therefore 0.1481 is the required probability here.