Question

A biased coin is thrown 100 times with probability of head = p and probability of tail = q. The probability of getting 55 heads in 100 trials is $\frac{100!}{55!\times 45!}{p}^{55}{q}^{45}$

• A. True
• B. False

Answer 1

The correct option is A True

We are throwing a die 100 times and we want to find the probability of getting 55 heads. Its given that probability of getting a head is p. How should we go about solving this problem? Should we write down all the favorable and possible outcomes and decide? It would take a lot of time to write down all the possible outcomes.
Let us consider one simple outcome where we get 55 heads.
HHHHH…..(55 heads in total)...TTTTT(45 tails)................(1)
Here, first 55 outcomes are heads and remaining outcomes are tails. Each of these trials are independent of any other trial.
So the probability of getting above outcome will be
$p\times p\times p\times p..\left(55times\right)..q\times q\times q\times q..\left(45times\right)=p^{55}{q}^{45}$
If we have 55 heads and 45 tails, the probability is going to be same $\left(p^{55}{q}^{45}\right)$ irrespective of where we place the ‘H’s and ‘T’s in (1). Now we have to find how many ways we can place 55 ‘H’s in those 100 places. This would be equal to ${}^{100}{C}_{55}=\frac{100!}{55!\times 45!}$
So the probability of getting 55 heads would be $\frac{100!}{55!\times 45!}{p}^{55}{q}^{45}$
$\Rightarrow$ The given statement is true