BTGmoderatorDC wrote:Out of 2 broken TV sets and 6 fixed TV sets in Joe's Fix-My-LCD store, 2 sets are selected randomly. What is the probability that at least 1 set is broken?
A. 11/28
B. 13/28
C. 15/28
D. 1/5
E. 1/8
OA B
Source: Economist Gmat
There are 8 TV's in total
2 are broken
6 are fixed
We want to find P(at least one TV is broken)
When it comes to probability questions involving at
least, it's often best to try using the
complement.
That is, P(Event A happening) = 1 - P(Event A
not happening)
So, here we get: P(getting at least 1 broken TV) = 1 -
P(not getting at least 1 broken TV)
What does it mean to
not get at least 1 broken TV? It means getting zero broken TVs.
So, we can write: P(getting at least 1 broken TV) = 1 -
P(getting zero broken TVs)
1 -
P(getting two FIXED TVs)
P(getting two FIXED TVs)
P(getting two FIXED TVs) = P(1st TV fixed
and 2nd TV is fixed)
= P(1st TV fixed)
x P(2nd TV is fixed)
= 6/8
x 5/7
= 30/56
=
15/28
So, P(getting at least 1 broken TV) = 1 -
P(not getting at least 1 broken TV)
= 1 -
15/28
= 13/28
Answer: B
Cheers,
Brent
