a) Given that exactly one head occurred, what is the probability that it occurred on the first toss.
I have this so far, and I just need help computing the numerator:
Total outcome $= 2^{10} = 1024$
Probability of $1$ head $= \frac{10}{1024} = 0.0098$
$$P(\text{first toss} \lor \text{one head}) = \frac{P(\text{first toss }\land \text{ one head})}{P(\text{one head})},$$
but I don't know how to compute $P(\text{first toss } \land \text{ one head})$
b) Given that the first two tosses were heads, what is the probability that there were exactly $3$ heads?
I don't have much for this one and I don't know how to compute the numerator or the denominator
Probability of $3$ heads $= \frac{C^{10}_3}{1024} = \frac{120}{1024} = 0.1172$
$$P(\text{3 heads }\lor \text{ first two toss are heads})= \frac{P(\text{3 heads }\land \text{ first two toss are heads})}{P(\text{first two toss are heads})}$$
$\endgroup$ 131 Answer
$\begingroup$If you're looking to use the conditional probability formula, you can.
There is one way to get exactly one head on the first toss of $10$: $HTTTTTTTTT.$ This is one out of $1024$ possible outcomes, so its probability is $1/1024.$
The probability of getting exactly one head (regardless of where) is $10/1024$.
Hence, the probability of getting a head on the first toss, given that you got only one head, is
$$P(A|B) = \frac{1/1024}{10/1024} = \frac{1}{10}.$$
For the second one: The probability of exactly three heads and two of the three falling on the first two tosses is $8/1024$ (just by counting them). The probability of getting the first two tosses come up heads is $1/4$. Hence the probability of getting exactly three heads, given the first two are heads, is
$$P(A|B) = \frac{8/1024}{1/4} = \frac{1}{32}.$$
$\endgroup$ 3
