A coin tossed 10 times.
what is the probability that the coin lands on 'Heads' less than 'Tails' given that the first toss was 'Heads'?
Say X indicating how many times the coin landed on Heads
Y indicating how many times the coin landed on Tails
B is the event where the first toss lands on 'Heads'.
I need to calculate $P(X<Y | B)$
from conditional probability I know that $P(X<Y | B) = \frac{P((X<Y) \bigcap B)}{P(B)}$
I know that $P(B) = 0.5$
and I think that $P((X<Y) \bigcap B)$ equals to $\sum_{i=0}^4 P((i<10-i) \bigcap B)$
but I'm stuck at this and would appriciate any help.
Best Answer
What you are looking for is that in 9 remaining coin tosses you have not more than three Heads. So $$ P(X \leq 3) = \sum_{i=0}^3 {9 \choose i} \left(\frac 12\right)^i \left(\frac 12\right)^{9-i} = \left(\frac 12\right)^9 \cdot \sum_{i=0}^3 {9 \choose i} \approx 0.25 $$