Question

There are 10 electric cars in a box and 3 are defective. Two electric cars are selected without replacement. What is the chance of selecting 1 good electric car and 1 defective electric car?

Use a tree diagram & remember independent & dependent concept.

A) .25

B) .233

C) .47

D) .21

Answer 1

Solution:

Given: There are 10 electric cars in a box and 3 are defective. Thus 10-3 = 7 good car.

Two electric cars are selected without replacement

Let G = Good car and D = Defective car

We have to find:

P( 1 good electric car and 1 defective electric car) = ............?

P( 1 good and 2 defective) = P(First Good & Second Defective) + P( First Defective and Second Good)

P( 1 good and 2 defective) = 7/10 X 3/9 + 3/10 X 7/9

P( 1 good and 2 defective) = 0.2333 +0.2333

P( 1 good and 2 defective) = 0.4666

P( 1 good and 2 defective) = 0.47