For an initial screening of an admission test, a candidate is given fifty problems to solve. If the probability that the candidate can solve any problem is $$\frac{4}{5}$$, then the probability that he is unable to solve less than two problems is

We have an admission-test screening in which a candidate attempts $$n=50$$ independent problems. For every single problem, the probability that the candidate solves it is given as $$\tfrac45$$. Consequently, the probability that the candidate does not solve (fails to solve) a problem is

$$q=1-\tfrac45=\tfrac15.$$

Let the random variable $$X$$ denote the number of problems the candidate is unable to solve out of the fifty. Because each problem is independent and has the same probability of failure $$q=\tfrac15$$, the distribution of $$X$$ is binomial with parameters $$n=50$$ and $$p=q=\tfrac15$$. The probability-mass function of a binomial random variable is stated first:

$$P(X=k)=\binom{n}{k}p^{\,k}(1-p)^{\,n-k}.$$

Here we require the probability that the candidate is unable to solve less than two problems, i.e.

$$P(X<2)=P(X=0)+P(X=1).$$

First term: $$P(X=0)$$. Substituting $$k=0$$ in the formula, we get

$$P(X=0)=\binom{50}{0}\Bigl(\tfrac15\Bigr)^{0}\Bigl(1-\tfrac15\Bigr)^{50} =1\cdot1\cdot\Bigl(\tfrac45\Bigr)^{50} =\Bigl(\tfrac45\Bigr)^{50}.$$

Second term: $$P(X=1)$$. Substituting $$k=1$$ gives

$$P(X=1)=\binom{50}{1}\Bigl(\tfrac15\Bigr)^{1}\Bigl(1-\tfrac15\Bigr)^{49} =50\cdot\tfrac15\cdot\Bigl(\tfrac45\Bigr)^{49} =10\Bigl(\tfrac45\Bigr)^{49}.$$

Adding the two probabilities. Factorising the common power $$\Bigl(\tfrac45\Bigr)^{49}$$ we have

$$P(X<2)=\Bigl(\tfrac45\Bigr)^{50}+10\Bigl(\tfrac45\Bigr)^{49} =\Bigl(\tfrac45\Bigr)^{49}\!\left[\tfrac45+10\right].$$

Convert the bracket to a single fraction:

$$\tfrac45+10=\tfrac45+\tfrac{50}{5}=\tfrac{4+50}{5}=\tfrac{54}{5}.$$

So,

$$P(X<2)=\frac{54}{5}\Bigl(\tfrac45\Bigr)^{49}.$$

This expression matches Option D in the given choices.

Hence, the correct answer is Option D.