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 $\frac{1}{4}$


  • 3
    $\begingroup$ Your answer seems correct $\endgroup$ – gen-z ready to perish Nov 15 '17 at 12:58
  • 1
    $\begingroup$ Do you mean to say the probability of 2 heads followed by 8 tails or are 1 or 0 heads in the first two tosses permissible as well.? $\endgroup$ – David Reed Nov 15 '17 at 13:04
  • $\begingroup$ Your answer is correct, because of the word "only" in the question. Their answer would be correct if the word "only" was omitted, i.e. if the remaining 8 coin tosses could have any outcomes instead of all tails. $\endgroup$ – Jaap Scherphuis Nov 15 '17 at 13:07
  • $\begingroup$ ... or if the word “only” came towards the end of the sentence @JaapScherphuis. $\endgroup$ – amd Nov 15 '17 at 16:07
  • $\begingroup$ @JaapScherphuis ,Chase,Jaap still the answer is not clear .is it $\frac{1}{1024}$ or $\frac{1}{4}$ $\endgroup$ – 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 $.


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