Probability of exactly one head from two coin toss (biased)

Two unfair coins are tossed with the probability of getting a tail equal to 0.25. What is the probability of getting exactly one head?

Attempt: I know that the possible combinations are TT,HH,HT,TH. So there are two combinations which give exactly one head. I know that the probability for an unbiased coin would be 2/4, but I am not sure how to do the calculation for an biased coin.

• I want to the know the probability of getting exactly one head not at least one head... – Sjoseph Jul 5 '17 at 12:57
• also the second unbiased at the bottom should say biased I think. just think about the probability of the arrangements the probability of two tails for example is $\frac{1}{4}\times\frac{1}{4}=\frac{1}{16}$ at last check. – user451844 Jul 5 '17 at 13:00

Well, since both coins are being tossed independently - by that it is meant that the result of the one does not affect in any way the result of the other, or in short "stochastic independence" - we can calculate the probability of each one of the cases that exactly one head appears by multiplication of the corresponding probabilities. This gives us:

1. $P(HT)=P(H)P(T)=0.75\times0.25=0.1875$,
2. $P(TH)=P(T)P(H)=0.25\times0.75=0.1875$,

where $HT$ means that the first coin gives heads and the second tails and $TH$ that the first coin gives tails and the second heads.

Now, since the eventualities of getting heads from the first coin and tails from the second and getting tails from the first coin and tails from the second are irrelevant, we can add the two possibilities, so:

$$P(\mbox{exactly one head appears})=0.1875+0.1875=0.375$$

Generally, the sample space of our problem is the set $$\Omega=\{HH,HT,TH,TT\}$$ and each of it's elements has a possibility that can be calculated as above, so, we have:

1. $P(HH)=0.75\times0.75=0.5625$,
2. $P(HT)=0.75\times0.25=0.1875$,
3. $P(TH)=0.25\times0.75=0.1875$,
4. $P(TT)=0.25\times0.25=0.0625$.

Note that the sum of all these probabilities is equal to 1, as expected.