What is the probability of getting two heads twice in $4$ tosses of two coins?
My Attempt:
No of trials $(n)=4$
Probability of success in one trial $(p)=\dfrac {1}{2}$
Probability of failure $(q)=\dfrac {1}{2}$. Thus, probability of getting two heads twice$=^{4}C_{2} (\dfrac {1}{2})^{2}. (\dfrac {1}{2})^{2}$ $$=\dfrac {3}{8}$$
The answer for this question is $\dfrac {27}{128}$. I couldn't get how's that.
What do you mean by "success"? It should be "rolling two heads in a toss of two coins". The probability of this is $$\frac12\times\frac12=\frac14. $$Then the probability of failure is $\frac34$. Using these numbers, you get $${4\choose2}\left(\frac14\right)^2\left(\frac34\right)^2=\frac{27}{128}$$as required.