[Math] A fair coin is tossed $\text{10 times}$. What is the probability that ONLY the first two tosses will yield heads

probabilityprobability theory

Question

A fair coin is tossed $\text{10 times}$. What is the probability that ONLY the first two tosses will yield heads?

My Solution

Here from the question we can conclude that the event i.e tossing $10$ coin are independent,hence

probablity of head=probablity of tail$$=\frac{1}{2}$$

Hence ,

$(\frac{1}{2})^2 \times (\frac{1}{2})^8 =(\frac{1}{2})^{10}=\frac{1}{1024}$

Am i correct ?Becuase answer given is $\frac{1}{4}$

Thanks

Best Answer

The event in the question can happen in only one way. The total number of possible outcomes, the sample space, is $ 2^{10}= 1024 $. So, the probability is $ 1/1024 $. If heads or tails were allowed in the remaining 8 tosses then the number of outcomes would be $ 2^8 = 256 $ and the probability would then be $ 256/1024 = 1/ 4 $.

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