We need to find On three consecutive flips of a coin, what is the probability that it will land on the same side each time?

Coin is tossed 3 times => Total number of cases = \(2^3\) = 8

Lets solve the problem using two methods

Method 1:

Out of the 8 cases there are only two cases in which all the outcomes are same. HHH and TTT.

=> Probability that all three produce the same result = \(\frac{2}{8}\) = \(\frac{1}{4}\)

So, Answer will be B

Method 2:

Let's find out the number of cases.
In the first toss we can get anything out of head or tail in 2 ways
In second toss we need to get the same value (head or tail) as we got in the first toss => 1 way
In third toss we need to get the same value (head or tail) as we got in the second toss => 1 way

=> Total number of ways = 2*1*1 = 2 ways

=> Probability that all three produce the same result = \(\frac{2}{8}\) = \(\frac{1}{4}\)

So, Answer will be B
Hope it helps!

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