1. In an examination, the probability of a candidate solving a question in $$ 1 / 2 $$. Out of given 5 questions in the exam, what is the probability that the candidate was able to answer at least 2 questions? A. $$ 1 / 64 $$ B. $$ 3 / 16 $$ C. $$ 1 / 2 $$ D. $$ 13 / 16 $$

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To find the probability that the candidate was able to answer at least 2 questions out of 5, we can use the binomial probability formula.

The probability of a candidate solving a question is $$ \frac{1}{2} $$, and the probability of not solving a question is also $$ \frac{1}{2} $$.

Let's denote the probability of solving at least 2 questions as $$ P $$.

We can calculate $$ P $$ using the binomial probability formula:

$$ P = \sum_{k=2}^{5} \binom{5}{k} \left(\frac{1}{2}\right)^k \left(\frac{1}{2}\right)^{5-k} $$

where $$ \binom{5}{k} $$ represents the number of ways to choose $$ k $$ questions out of 5.

Now, let's calculate the probability using the formula.
Find the sum by following steps:
- step0: Solution:
  $$\sum _{k=2}^{5} { }_{5}C_{k}\left(\frac{1}{2}\right)^{k}\left(\frac{1}{2}\right)^{5-k}$$
- step1: Expand the expression:
  $$\sum _{k=2}^{5}\frac{5!}{k!\times \left(5-k\right)!}\times \left(\frac{1}{2}\right)^{k}\left(\frac{1}{2}\right)^{5-k}$$
- step2: Calculate the value:
  $$\sum _{k=2}^{5}\frac{120}{k!\times \left(5-k\right)!}\times \left(\frac{1}{2}\right)^{k}\left(\frac{1}{2}\right)^{5-k}$$
- step3: Multiply:
  $$\sum _{k=2}^{5}\frac{15}{2^{2}\times k!\times \left(5-k\right)!}$$
- step4: Rewrite the expression:
  $$\sum _{k=2}^{5}2^{-2}\left(k!\right)^{-1}\left(\left(5-k\right)!\right)^{-1}\times 15$$
- step5: Convert the expressions:
  $$2^{-2}\left(2!\right)^{-1}\left(\left(5-2\right)!\right)^{-1}\times 15+2^{-2}\left(3!\right)^{-1}\left(\left(5-3\right)!\right)^{-1}\times 15+2^{-2}\left(4!\right)^{-1}\left(\left(5-4\right)!\right)^{-1}\times 15+2^{-2}\left(5!\right)^{-1}\left(\left(5-5\right)!\right)^{-1}\times 15$$
- step6: Calculate:
  $$\frac{5}{16}+\frac{5}{16}+\frac{5}{32}+\frac{1}{32}$$
- step7: Calculate:
  $$\frac{13}{16}$$
The probability that the candidate was able to answer at least 2 questions out of 5 is $$ \frac{13}{16} $$ or 0.8125.

Therefore, the correct answer is D. $$ \frac{13}{16} $$.
The probability that the candidate answered at least 2 questions is $$ \frac{13}{16} $$.

In an examination, the probability of a candidate solving a question in . Out of given 5 questions in the exam, what is the probability that the candidate was able to answer at least 2 questions?
A.
B.
C.
D.

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