Two unbiased coins are tossed once, what is the probability of getting exactly one head

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.

Which of the following arguments are correct and which are not correct ? Give reasons for your answer.

If a die is thrown once, the possible outcomes areS = {1, 2, 3, 4,5, 6} i.e.    n(S) = 6Let A be the favourable outcomes of getting add number. ThenA = {1,3,5} i.e.    n(S) = 3

Therefore, P(A) =