Question

A fair coin is tossed 9 times.

(A) What is the probability of tossing a tail on the 9th toss, given that the preceding 8 tosses were heads?

(B) What is the probability of getting either 9 heads or 9 tails?

(A) What is the probability of tossing a tail on the 9th toss, given that the preceding 8 tosses were heads?

(B) What is the probability of getting either 9 heads or 9 tails?

Answer 1

(a) As each coin toss is an independent event so the probability of getting a tail on 9th toss will be 1/2 as there's a 50% chance of getting a tail.

The next part is different because we are looking at cumulative probability. Also, streaks like getting 3 heads in a row are related to each other so we cannot consider them as independent events.

(b)

We will use Binomial Distribution in this case.

Since the coin is tossed 9 times. N which is the number of trails will be 9.

P = Probability of heads = 0.5

Q = Probability of tails = 0.5

X = No. heads occurring in each toss.

So, now we have to find the probability for exactly 9 heads or 9 tails.

P(x = 9) = C(9, 9)P^x Q^N-x

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0 9C9*0.5°x0.5 1 512

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