Question

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?

Answer 1

`5/16` See explanation

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.