[email protected] wrote:Q) A drawer has 2 identical red socks, 4 identical black socks and 6 identical white socks. What is the fewest number of socks that must be pulled out to have at least a 50% chance of pulling out at least one red socks?
A) 1
B) 2
C) 3
D) 4
E) 10
So, we have 2 red socks and 10 non-red socks.
To begin, probability questions are great for applying some intuition.
Consider answer choice A. We know that if we draw 1 sock, the probability of selecting a red sock is less than 0.5 (50%) So, eliminate A.
Now consider answer choice E. If we draw 10 socks, the probability of selecting at least one red sock is VERY VERY high. Much greater than 0.5 We can apply some intuition here and conclude that we must be able to reach a probability of 0.5 with few than 10 draws. So, eliminate E.
This leaves us with drawing 2, 3 or 4 socks.
Let's try 3 (answer choice C) and go from there.
If we draw 3 socks, what is P(draw at least 1 red sock)?
When it comes to probability questions involving "at least," it's often a good idea to use the complement.
That is, P(Event A happening) = 1 - P(Event A
not happening)
So, here we get: P(getting at least 1 red sock) = 1 -
P(not getting at least 1 red sock)
In other words, P(getting at least 1 red sock) = 1 -
P(getting zero red socks)
P(getting zero red socks) = P(no red sock on 1st draw
AND no red sock on 2nd draw
AND no red sock on 3rd draw)
= P(no red sock on 1st draw)
x P(no red sock on 2nd draw)
x P(no red sock on 3rd draw)
= (10/12)
x (9/11)
x (8/10)
=
6/11
So, when we select 3 socks, P(at least 1 red sock) = 1 -
6/11
= 5/11 (less than 0.5)
Since we want P(at least 1 red sock) to be 0.5 or greater,
we must select more than 3 socks.
IMPORTANT: At this point, the answer must be
D, so I'd select
D without performing any additional calculations.
Cheers,
Brent
