A coin is tossed 5 times. Find P(at most 4 tails)
a. 3/16
b.13/16
c.1/32
d.31/32
So far I got:
There are two possible outcomes for each toss, so there are 32 outcomes. Four at the most must be tails, so s=4. The other 28 outcomes are failures so f=28.
P(at most 4 tails)= (s)/(s+f)
(4)/(4+28) = 4/32 = 1/8
but as you see that is not a choice can someone explain what I am doing wrong.
a. 3/16
b.13/16
c.1/32
d.31/32
So far I got:
There are two possible outcomes for each toss, so there are 32 outcomes. Four at the most must be tails, so s=4. The other 28 outcomes are failures so f=28.
P(at most 4 tails)= (s)/(s+f)
(4)/(4+28) = 4/32 = 1/8
but as you see that is not a choice can someone explain what I am doing wrong.
