A coin is tossed 4 times, what is the probability of Not getting 2 heads would mean getting all tails or 3 tails and 1 head?

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Answer 1 · 2017-06-08T08:44:07

#5/16# See explanation

Let me assign p as getting head and q as getting tail.
#p+q=1#. And #p=1/2# (for a fair coin). Also #q=1/2# (for a fair coin).

Now compute getting all tails (four trials independent)

Probability of having all tails = (1/2)(1/2)(1/2)*(1/2)=1/16# (four trials and all ended with tail)

Probability of one head (the first trial) and three tails (after first trial):

#=(1/2)*(1/2)*(1/2)*(1/2)=1/16#

Probability of one head (the second trial) and three tails (the first, the third, and the final trials):

#=(1/2)*(1/2)*(1/2)*(1/2)=1/16#

Probability of one head (the third trial) and three tails (the first, the second, and the final trials):

#=(1/2)*(1/2)*(1/2)*(1/2)=1/16#

Finally, probability of one head (the final trial) and three tails (the first, the second, and the third trials):

#=(1/2)*(1/2)*(1/2)*(1/2)=1/16#

Total probability showing only one (or none) head and at least three tails (order not important)
P#=(1/16)+(1/16)+(1/16)+(1/16)+(1/16)=5/16#

which is #31.25%# probability.