# Thread: Prob that person A tossed coin once to get H and that the other 1 needed 9 throws

1. ## Prob that person A tossed coin once to get H and that the other 1 needed 9 throws

Hey, I have the following question... I know the probabilities for each event happening independently, but not in the way the question is asking. Should I multiply or add the probabilities? What should I do? Cheers!

The problem says that person A tosses a fair coin until it lands heads, and that then person B does the same. In the end, they toss it 10 times. What is the probability that person A only tossed it once?

Well, I have that the probability that person A gets heads with only once throw is 0.5, and that the probability that person gets heads with 9 throws is 9/(2^9)=9/512=0.02

So I don't know if I should add them (which would give P=0.52) or multiply them... or something different. Thanks a lot!

2. ## Re: Prob that person A tossed coin once to get H and that the other 1 needed 9 throws

Hello, buenogilabert!

Person $A$ tosses a fair coin until it lands heads, then person $B$ does the same.
In the end, they toss it 10 times.
What is the probability that person $A$ tossed it only once?

$A$ tossed once and got a Head,
. . then $B$ tossed nine times to grt a Head.

We want: . $P(\text{A: Head on 1st toss}) \;\text{ and }\;P(\text{B: 8 Tails, then a Head})$

. . . . . . . . . . . . . . . . $\frac{1}{2} \qquad\qquad\quad\; \times \qquad\qquad\quad \left(\frac{1}{2}\right)^8\cdot\frac{1}{2}$

Therefore: . $\text{Probability} \:=\:\left(\frac{1}{2}\right)^{10} \:=\:\frac{1}{1024}$