Question

A problem in statistics is given to 3 students A, B and C. Their chances of solving problems are
$\dfrac{1}{3},\,\,\dfrac{1}{4}\,\,and\,\,\dfrac{1}{5}$ respectively. If all the students independently what is the
probabilities that:
a) Problem is solved
b) Problem is not solved
c) Exactly 2 students solve the problem

Answer 1

Hint: use a formula of probability when we solve the part of this question.
Formula used: \[probability{\text{ }} = \dfrac{{favourable{\text{ }}outcomes}}{{total\,\,number\,\,of\,\,outcomes}}\]

Complete step-by-step answer:
Consider the following situation:

Probability of A solving the question
\[probability{\text{ }} = \dfrac{{favourable{\text{ }}outcomes}}{{total\,\,number\,\,of\,\,outcomes}}\]
$P(A) = \dfrac{1}{3}$ ….(1)
Probability of A not solving the question
$ \Rightarrow P(\bar A) = 1 - \dfrac{1}{3} = \dfrac{2}{3}$ …..(2)
Similarly, P(B) =$\dfrac{1}{4}\,\,and\,\,$ …..(3)
$P(\bar B) = 1 - \dfrac{1}{4} = \dfrac{3}{4}$ ….(4)
\[probability{\text{ }} = \dfrac{{favourable{\text{ }}outcomes}}{{total\,\,number\,\,of\,\,outcomes}}\]
$P(C) = \dfrac{1}{5}\,\,and\,\,$ ….(5)
$P(\bar C) = 1 - \dfrac{1}{5} = \dfrac{4}{5}$ ….(6)
(a) Probability that question is solved = 1- probability that question is not solved.
$ \Rightarrow P(\bar Q)$represents,
Case 8 of the table 1.1
\[P(\bar Q) = P(\bar A) \times P(\bar B) \times P(\bar C)\]
From equation (2), (4) and (6)
$ = \dfrac{2}{3} \times \dfrac{3}{4} \times \dfrac{4}{5} = \dfrac{2}{5}$
Also, $P(\bar Q) = 1 - \dfrac{2}{5} = \dfrac{3}{5}$
Hence, probability of showing the question is $\dfrac{3}{5}$
(b) From (a) P($\bar Q$) =$\dfrac{2}{5}$
Probability of not solving the question $ = \dfrac{2}{5}$
(c) Probability that exactly two students show the question are the case 2, 3 and 5 from table 1.1 = P(exact two).

P(exact two) = case 2 or case 3 or case 5
$ = P(A).P(B).P(\bar C) + P(A).P(\bar B).P(C) + P(\bar A).P(B).P(C)$
From equation (1), (2), (3), (4), (5) and (6) we will get,
$ = \dfrac{1}{3} \times \dfrac{1}{4} \times \dfrac{4}{5} + \dfrac{1}{3} \times \dfrac{3}{4} \times \dfrac{1}{5} +
\dfrac{2}{3} \times \dfrac{1}{4} \times \dfrac{1}{5}$
$
= \dfrac{{4 + 3 + 2}}{{3 \times 4 \times 5}} \\
= \dfrac{9}{{3 \times 4 \times 5}} \\
= \dfrac{3}{{20}} \\
$
Probability that exactly two students solve the probability is $\dfrac{3}{{20}}$.

Note: When solving problems like these, make sure to find the complement of the probability(that is probability of the event not occurring) and then subtract it from 1 to get the required probability.