swerve wrote:The probability is 1/2 that a certain coin will turn up heads on any given toss. If the coin is to be tossed four times, what is the probability that on at least one of the tosses the coin will turn up tails?
A. 1/16
B. 1/8
C. 1/2
D. 7/8
E. 15/16
When it comes to probability questions involving "at least," it's 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 tails) = 1 -
P(NOT getting at least 1 tails)
What does it mean to NOT get at least 1 tails? It means getting zero tails.
So, we can write: P(getting at least 1 tails) = 1 -
P(getting zero tails)
Now let's calculate
P(getting zero tails)
What needs to happen in order to get zero tails?
Well, we need heads on the 1st toss
and heads on the 2nd toss
and heads on the 3rd toss
and heads on the 4th toss
We can write
P(getting zero tails) = P(heads on 1st
AND heads on 2nd
AND heads on 3rd
AND heads on 4th)
This means that
P(getting zero tails) = P(heads on 1st)
x P(heads on 2nd)
x P(heads on 3rd)
x P(heads on 4th)
Which means
P(getting zero tails) = (1/2)
x(1/2)
x(1/2)
x(1/2) =
1/16
We're now ready to answer the question.
P(getting at least 1 tails) = 1 -
P(NOT getting at least 1 tails)
= 1 -
1/16
= 15/16
= E
Cheers,
Brent
