# What is the probability that your 10-toss sequence is either all heads or all tails?You toss a balanced coin 10 times and write down the resulting sequence of heads and tails, such as HTTTHHTHHH.

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Interpreting a balanced coin to be an unbiased one , the probability of showing head or tail in a single toss is (1/2) each.Or P(H)=P(T) =(1/2).

If A and B are two independent events, then P(AB) = P(A)P(B).

Therefore, probability of getting 10 heads in 10 tosses (tosses being independent) = P(HHHHHHHHHH) = (1/2)(1/2)... 10 factors=(1/2)^10.

Similarly Probability of getting 10 tails= P(TTTTTTTTTT)=(1/2)^10.

Since both events are exclusive, the probability of getting 10 heads or 10 tails is equal to the sum of the the individual probabilities = P(HHHHHHHHHH)+P(TTTTTTTTTT) = (1/2)^10+(1/2)^10 = 2(1/2)^10 = 2^(-9) =1/512 = 1.953125*10^(-3)

The probability that a single toss will be head only is 0.5.

The probability that in a sequence if n number of tosses all results will be heads is given by the expression:

Probability of all of n tosses resulting in heads = 0.5^n

Therefor the probability of all of 10 tosses resulting in heads = 0.5^10

= 0.0009765625

Similarly probability of all the 10 tosses being is = 0.0009765625

And the probability of all 10 tosses being heads or all tosses being tails is sum of probabilities of these two possibilities.

Therefor probability of all 10 tosses being heads or all tosses being tails

= 0.0009765625 + 0.0009765625 = 0.001953125