tejal777 wrote:
The probability is 0.6 that an “unfair” coin will turn up tails on any given toss. If the coin is tossed 3 times, what is the probability that at least one of the tosses will turn up tails?
A. 0.064
B. 0.36
C. 0.64
D. 0.784
E. 0.936
Where's my error:
P(1)+P(2)+P(3)
6/10.4/10.4/10 + 6/10.6/10.4/10 +6/10.6/10.6/10
=0.516
ASIDE---------------
We want P(select at least 1 tails)
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)-------------------
P(getting zero tails)P(getting zero tails) = P(heads on 1st toss
AND heads on 2nd toss
AND heads on 3rd toss)
= P(heads on 1st toss)
x P(heads on 2nd toss)
x P(heads on 3rd toss)
= 0.4
x 0.4
x 0.4
=
0.064So, P(getting at least 1 tails) = 1 -
0.064 = 0.936
Answer: E
Cheers,
Brent
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