A fair coin is tossed 3 times.
Given the first tossing is a head, what is the probability to have three heads?
If you have been told that at least two of the tosses are heads, what is the probability to have a head in the first toss?
If someone could explain how to do this to me using a visualisation method (as in writing down the possible outcomes then deducing from there) I would gladly appreciate it.
Is question 1 just 0.5 x 0.5 = 0.25 since the results of the next 2 tosses are independent of the first?
For question 2, P(2 heads)= 4/8 and P(1st toss heads)=3/8 (given that there must be 2 heads), do I multiply them together?
Thanks in advance.
1) If the first tossing is a head then there are $4$ possibilities: HHH, HTH, HHT, HTT. They have equal probability to occur and in exactly $1$ of the possibilities (HHH) you have three heads. This amounts in a probability of $\frac{1}{4}$.
2) If at least two of the tossings are heads then there are $4$ possibilities: HHH, HHT, HTH and THH. They have equal probability to occur and in exactly $3$ of the possibilities (HHH, HHT and HTH) the first toss is a head. This amounts in a probability of $\frac{3}{4}$.