Question

Printed circuit cards are placed in a functional test after being populated with semiconductor chips. Suppose that there are 8 blue credit cards, 6 red credit cards and 4 blue debit cards in a box. It is known that color and card type are to be independent when a card is selected at random from this box.

a) How many red debit cards must be present?

b) Suppose that, 5 cards are randomly selected in a row with replacement. What is the probability that 2 blue credit cards and 2 blue debit cards are selected?

c) Suppose that, 5 cards are randomly selected in a row without replacement. What is the probability that 2 blue credit cards and 2 blue debit cards are selected?

Answer 1

a.

let x = no. of red debit card

for independent :

P(red | debit card) = P(red | credit card)

no. of red debit card / no. of debit card = no. of red credit card / no. of credit card

x/(4+x) = 6/(6+8)

x*(6+8) = 6*(4+x)

14*x = 24 + 6*x

8*x = 24

x = 3

therefore : 3 red debit cards

b.

total cards = 8+6+4+3 = 21

P(blue credit card) = 8/21

P(blue debit card) = 4/21

probability that 2 blue credit cards and 2 blue debit cards are selected

= P(2 blue credit card out of 5 cards)*P(2 blue debit cards out of 3 remaining cards)

=[ 5C2*(8/21)^2 * (1 - 8/21)^3 ] * [ 3C2*(4/21)^2 * (1 - 4/21)^1 ]

= [ 10*(8/21)^2 * (1 - 8/21)^3 ] * [ 3*(4/21)^2 * (1 - 4/21)^1 ]

= 0.0303

c.

without replacement :

no. of (blue credit card) = 8

no. of (blue debit card) = 4

no. of other cardss = 21-8-4 = 9

P(2 blue credit cards and 2 blue debit cards) = [(select 2 blue credit card from 8)*(select 2 blue debit card from 4)*(select 1 other cards from 9 other card)]/(select 5 cards from 21 cards)

= [8C2 * 4C2 * 9C1] / (21C5)

= [28*6*9 ] / 20349

= 0.0743

(please UPVOTE)