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

## 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

• @JaapScherphuis ,Chase,Jaap still the answer is not clear .is it $\frac{1}{1024}$ or $\frac{1}{4}$ – laura Nov 16 '17 at 5:21
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 $256/1024 = 1/ 4$.