Question

Out of 100 students two sections of 40 and 60 are formed. If you and your friend are among the 100 students, what is the probability that
(a). you both enter the same sections?
(b). you both enter the different sections?

Answer 1

Hint: This is a case of conditional probability. It has two conditions one depending on other where firstly 40 or 60 students are chosen from 100 students and secondly from those 40 or 60 students 2 students are chosen. Then the condition indicates a SUM/OR statement which gives us the final result. The second statement refers to the not fulfilling the first statement.

Complete step-by-step answer:

For the first condition states that both enter the same section, means either they enter section A or section B.
We know the formula,
$P\left( E \right)=\dfrac{n\left( E \right)}{n\left( S \right)}...........(i)$
Here, P(E) is the probability of an event or events which is asked, n(E) is the number of favourable events and n(S) is the total number of all the events that can occur possibly.
In the given question there are a total of 100 students and out of these we are concerned about two students.
As we know that number of ways of selecting ‘r’ items from n items is given as:
$^{n}{{C}_{r}}$
So, the total number of all possible outcomes for selecting two students out of 100 students is,
$^{100}{{C}_{2}}..........(ii)$
This becomes our n(S) for the whole question.
Now let us consider case (a): We both enter the same section, i.e., the two students enter the same section.
If they enter section ‘A’ means they are among 40 students chosen from 100 students and out of those 40 they are those 2 who are in section ‘A’.
So, the number of ways of selecting these 2 students out of 40 students is,
$n(A){{=}^{40}}{{C}_{2}}...........(iii)$
We can define the probability of selecting 2 students out of 40 students from a total of 100 students from equation (i) as,
$P\left( A \right)=\dfrac{n\left( A \right)}{n\left( S \right)}$
Substituting values from equation (ii) and (iii), we get
$P(A)=\dfrac{^{40}{{C}_{2}}}{^{100}{{C}_{2}}}$
Now we know, $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$, applying this formula in above equation, we get
$P(A)=\dfrac{\dfrac{40!}{2!\left( 40-2 \right)!}}{\dfrac{100!}{2!\left( 100-2 \right)!}}$
$P(A)=\dfrac{\dfrac{40!}{2!38!}}{\dfrac{100!}{2!98!}}$
$P(A)=\dfrac{\dfrac{40\times 39\times 38!}{2\times 1\times 38!}}{\dfrac{100\times 99\times 98!}{2\times 1\times 98!}}$
$P(A)=\dfrac{40\times 39}{100\times 99}$
$P(A)=\dfrac{156}{990}$
If they enter section B means they are among 60 students chosen from 100 students and out of those 60 they are those 2 who are in section B.
So, the number of ways of selecting these 2 students out of 60 students is,
$n(B){{=}^{60}}{{C}_{2}}...........(iv)$
We can define the probability of selecting 2 students out of 60 students from a total of 100 students using equation (i) as,
$P\left( B \right)=\dfrac{n\left( B \right)}{n\left( S \right)}$
Substituting values from equation (ii) and (iv), we get
$P(B)=\dfrac{^{60}{{C}_{2}}}{^{100}{{C}_{2}}}$
Now we know, $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$, applying this formula in above equation, we get
$P(B)=\dfrac{\dfrac{60!}{2!\left( 60-2 \right)!}}{\dfrac{100!}{2!\left( 100-2 \right)!}}$
$P(B)=\dfrac{\dfrac{60!}{2!58!}}{\dfrac{100!}{2!98!}}$
$P(B)=\dfrac{\dfrac{60\times 59\times 58!}{2\times 1\times 58!}}{\dfrac{100\times 99\times 98!}{2\times 1\times 98!}}$
$P(B)=\dfrac{60\times 59}{100\times 99}$
$P(B)=\dfrac{354}{990}$
Therefore, the total probability of both entering same section is
$\begin{aligned} & P(A)+P(B)=\dfrac{156}{990}+\dfrac{354}{990} \\\ & P(A)+P(B)=\dfrac{156+354}{990} \\\ & P(A)+P(B)=\dfrac{510}{990} \\\ & P(A)+P(B)=\dfrac{51}{99}=\dfrac{17}{33} \\\ \end{aligned}$

Now let us consider case (b): We both enter the different sections, i.e., the two students enter different sections.
Now for second condition to entering in different sections the probability will be,
Probability of entering in different sections = 1 – Probability of entering in same sections
$1-[P(A)+P(B)]=1-\dfrac{17}{33}$
$\begin{aligned} & =\dfrac{33-17}{33} \\\ & =\dfrac{16}{33} \\\ \end{aligned}$

Hence, the probability that
(a). you both enter the same sections is $\dfrac{17}{33}$
(b). you both enter the different sections is $\dfrac{16}{33}$

Note: The chances of making mistakes are in cases when applying conditional probability. If the conditional probabilities have been found correctly then caution should be taken when using the OR/SUM conditions.
One can go wrong while selecting ‘r’ items from ‘n’ items. He or she may apply $P_{r}^{n}$ identity in place of $^{n}{{C}_{r}}$ .