Question

3. (10 points) Peter wants to draw a birthday card with pencil on paper. But, he cannot draw without making mistakes. He needs a pencil that has a sharp tip and a pencil that has a clean eraser (the good tip and good eraser need not be on the same pencil). The problem is Peter’s little sister Suzy. She often gets into Peter’s pencil box, breaks the tips and destroys the erasers. In fact, Peter figures any pencil has only a 10% chance of having a good tip and a 20% chance of a good eraser. There are 8 pencils.

1. What is the chance of Peter being set to draw the card?

2. What if he had 20 pencils? (you lose 2 points if you do this “the long way”

Answer 1

2) Peter just wants a pencil with good tip and a pencil with clean eraser. Let per no. of pencils with good tip and Eino. of pencils have clean eraser, then Be~ Binomial (8,0.1) and En Binomial (8,0.2) Now, card can be drawn if eal and E), hence P(card can be drawn) P(621, E3,1) = PC1821) PC E21) =[-P(R=0}] [1-PCE =o)] = (0.5695)*(0.8322) = 0.074 b) Similarly here to - Pe ~ Binomial (20,0.1) En Binomial (20,0.2) hence, P(card can be drawn) = P( Pe>), ESI) = [-PLV2501] [1-PC6 -01] 0.8784 x 0.9885 ī0.8683