TS Inter 2nd Year Maths 2A Solutions Chapter 9 Probability Ex 9(b)

Students must practice this TS Intermediate Maths 2A Solutions Chapter 9 Probability Ex 9(b) to find a better approach to solving the problems.

TS Inter 2nd Year Maths 2A Solutions Chapter 9 Probability Ex 9(b)

I.
Question 1.
If 4 fair coins are tossed simultaneously, then find the probability that 2 heads and 2 tails appear.
Solution:
If getting Head and Tail are denoted by H, T the number of sample points in the sample space n(S ) = 2⁴ = 16.
If E is the event of getting 2 heads and 2 tails then E = {THHT, TTHH, HTHT, HHTT, TTHH, HTTH}
∴ n ( E ) = 6
∴ Probability of getting 2 heads and 2 tails.
P(E) = \(\frac{n(E)}{n(S)}=\frac{6}{16}=\frac{3}{8}\)

Question 2.
Find the probability that a non-leap year contains
(i) 53 Sundays
(ii) 52 Sundays only.
Solution:
A non-leap year contains 365 days in which there are 52 full weeks and one day.
52 full weeks contain 52 × 7 = 364 days and
left out of one day may be either Sun, Mon, Tues, Wed, Thu, Fri or Sat.
∴ n(S) = 7.

i) For a non-leap year to contain 53 Sundays, we have only on possibility to have 365th day a Sunday.
∴ n(E) = 1

ii) For a non-leap year to contain 52 Sundays, we have 6 possibilities for 365th day be a day other Sunday
∴ n ( E ) = 6
∴ Probability of getting 52 Sundays in a non leap year is
∴ p(E) = \(\frac{\mathrm{n}(\mathrm{E})}{\mathrm{n}(\mathrm{S})}=\frac{6}{7}\).

Question 3.
Two dice are rolled. What is the probability that none of the dice shows the number 2?
Solution:
n = 6, m = 2 i.e., n(S) = 6, n (E) = 2
P ( E ) = Probability of not getting 2 when a single die is rolled = \(\frac{5}{6}\).
∴ The probability of not getting 2 when two dice are rolled = \(\left(\frac{5}{6}\right) \times\left(\frac{5}{6}\right)=\left(\frac{5}{6}\right)^2\).

Question 4.
In an experiment of drawing a card at random from a pack, the event of getting a spade is denoted by A and getting a pictured card (King, Queen or Jack) is denoted by B. Find the probabilities of A, B, A ∩ B and A ∪ B.
Solution:
Total number of cards in the pack n ( S ) = 52
If A is the event of getting a spade card then n (A) = 13
∴ Probability of getting a spade card,
P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{13}{52}=\frac{1}{4}\).

If B is the event of getting a picture card (King, Queen or Jack) then n (B) = 12
(there are 4 kings, 4 queens and 4 jacks)
∴ Probability of getting a picture card = \(\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{12}{52}=\frac{3}{13}\)
∴ P(B) = \(\frac{3}{13}\)
Common to the events A and B there is a spade king, spade queen and spade jack
∴ n(A ∩ B) = 3
∴ Probability of getting (A ∩ B) is
P(A ∩ B) = \(\frac{\mathrm{n}(\mathrm{A} \cap \mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{3}{52}\)
By addition theorem,
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= \(\frac{13}{52}+\frac{12}{52}-\frac{3}{52}=\frac{22}{52}=\frac{11}{26}\).

Question 5.
In a class of 60 boys and 20 girls, half of the boys and half of the girls know cricket. Find the probability of the event that a person selected from the class is either a boy or a girl who knows cricket.
Solution:
Let E, be the event of selecting a boy and E, be the event of selecting a person who knows cricket.
P(E₁) = \(\frac{60}{80}\);
P(E₂) = \(\frac{40}{80}\)
(30 boys + 10 girls who knows cricket)
P (E₁ ∩ E₂) = \(\frac{30}{80}\)
(∵ 30 boys are common to E₁ and E₂)
∴ Probability of selecting a boy or a girl who knows cricket is .
P(E₁ ∪ E₂) = P(E₁) + P(E₂) – P(E₁ ∩ E₂)
= \(\frac{60}{80}+\frac{40}{80}-\frac{30}{80}=\frac{70}{80}=\frac{7}{8}\)
∴ ProbabilIty of the event that a person selected from the class is either a boy or a girl who knows cricket = \(\frac{7}{8}\).

Question 6.
For any two events A and B, show that P\(\left(A^c \cap B^c\right)\) = 1 + P(A ∩ B) – P(A) – P(B)
Solution:
\(\left(A^c \cap B^c\right)\) = \((A \cup B)^c\)
∴ P\(\left(A^c \cap B^c\right)\) = P\((A \cup B)^c\)
= 1 – P(A ∪ B) (∵ P(S) = 1)
= 1 – [P (A) + P (B ) – P (A ∩ B)]
= 1 + P(A ∩ B) – P(A) – P(B)
[From Demorgan’s law \(\left(A^c \cap B^c\right)\) = \((A \cup B)^c\)].

Question 7.
Two persons A and B are rolling a dice on the condition that the person who gets 3 will win the game. If A starts the game, then find the probabilities of A and B respectively to win the game.
Solution:
Two persons A and B toss a die.
The probability of throwing 3 with a die = \(\frac{1}{6}\)
The probafility of not throwing 3 with the die = 1 – \(\frac{1}{6}\) = \(\frac{5}{6}\)
If A is to win, A must throw 3 in the 1st or 2nd or 3rd rounds etc.
P(A) = \(\frac{1}{6}+\frac{5}{6} \cdot \frac{5}{6} \cdot \frac{1}{6}+\frac{5}{6} \cdot \frac{5}{6} \cdot \frac{5}{6} \cdot \frac{5}{6} \cdot \frac{1}{6}+\ldots .\)
which is an infinite G.P. with a = \(\frac{1}{6}\)
r = \(\frac{5}{6} \cdot \frac{5}{6}=\left(\frac{5}{6}\right)^2\)
∴ P(A) = S_∞
= \(\frac{a}{1-r}=\frac{\frac{1}{6}}{1-\left(\frac{5}{6}\right)^2}=\frac{6}{11}\)
B wins when A fails to win
∴ P(B) = \(P(A)^c\)
= 1 – P(A)
= 1 – \(\frac{6}{11}\) = \(\frac{5}{11}\).

Question 8.
A, B, C are 3 newspapers from a city, 20% of the population read A, 16% read B, 14% read C, 8% both A and B, 5% both A and C, 4% both B and C and 2% all the three. Find the percentage of the population who read atleast one newspaper.
Solution:
Given P(A) = \(\frac{20}{100}\),
P(B) = \(\frac{16}{100}\)
P(C) = \(\frac{14}{100}\)
P(A ∩ B) = \(\frac{8}{100}\)
P(A ∩ C) = \(\frac{5}{100}\)
P(B ∩ C) = \(\frac{4}{100}\) and
P(A ∩ B ∩ C) = \(\frac{2}{100}\)
∴ The percentage of the population who read atleast one newspaper.
∴ P(A ∪ B ∪ C) = P(A) + P(B) + P(C) – P(A ∩ B) – P(B ∩ C) – P(C ∩ A) + P(A ∩ B ∩ C)
= \(\frac{20}{100}+\frac{16}{100}+\frac{14}{100}-\frac{8}{100}-\frac{4}{100}-\frac{5}{100}+\frac{2}{100}\)
= \(\frac{35}{100}\)
∴ 35 % read as atleast one newspaper.

Question 9.
If one ticket is randomly selected from tickets numbered 1 to 30, then find the probability that the number on the ticket is
i) a multiple of 5 or 7
ii) a multiple of 3 or 5.
Solution:
Given n (S) = 30
i) Let E₁ he the event of getting a number which is a multiple of 5.
Let E₂ be the event of getting a number which is a multiple of 7..
n(E₁) = {5, 10, 15, 20, 25, 30} = 6
and n(E₂) = {7, 14, 21, 28} = 4
∴ P(E₁) = \(\frac{n\left(E_1\right)}{n(S)}=\frac{6}{30}\)
P(E₂) = \(\frac{\mathrm{n}\left(\mathrm{E}_2\right)}{\mathrm{n}(\mathrm{S})}=\frac{4}{30}\)
∴ P(E₁ ∪ E₂) = P(E₁) + P(E₂)
(∵ P(E₁ ∩ E₂) = 0, because E₁ ∩ E₂ = Φ)
= \(\frac{6}{30}+\frac{4}{30}=\frac{10}{30}=\frac{1}{3}\)

ii) Let E₃ be the event of getting ticket which is a number multiple of 3.
Let E₄ be the event of getting a number which is a multiple of 5.
n(E₃) = {3, 6, 9, 12, 15, 18, 21, 24, 27, 30}
n(E₃) = 10
∴ P(E₃) = \(\frac{n\left(E_3\right)}{n(S)}=\frac{10}{30}=\frac{1}{3}\)
E₄ = {5, 10, 15, 20, 25, 30} = 6
n(E₄) = 6,
∴ P(E₄) = \(\frac{\mathrm{n}\left(\mathrm{E}_4\right)}{\mathrm{n}(\mathrm{S})}=\frac{6}{30}=\frac{1}{5}\)
n(E₃ ∩ E₄) = {15, 30} = 2
∴ P(E₃ ∩ E₄) = \(\frac{\mathrm{n}\left(\mathrm{E}_3 \cap \mathrm{E}_4\right)}{\mathrm{n}(\mathrm{S})}=\frac{2}{30}\)
∴ P(E₃ ∪ E₄) = P(E₃) + P(E₄) – P(E₃ ∩ E₄)
= \(\frac{10}{30}+\frac{6}{30}-\frac{2}{30}=\frac{14}{30}=\frac{7}{15}\)
∴ Probability for the number on the ticket to be a multiple of 3 or 5 is \(\frac{7}{15}\).

Question 10.
If two numbers are selected randomly from 20 consecutIve natural numbers, find the probability that the sum of the two numbers is
i) an even number
ii) an odd number.
Solution:
i) Let A be the event that the sum of the numbers is even when two numbers chosen out of 20 consecutive positive integers.
n(S) = ²⁰C₂ = 190
n (A) = ¹⁰C₂ + ¹⁰C₂
= 45 + 45 = 90
∴ Probability for the sum of two numbers to be even = \(\frac{90}{190}=\frac{9}{19}\)

ii) Probability for the sum of numbers to be odd = 1 – \(\frac{9}{19}\) = \(\frac{10}{19}\).

Question 11.
A game consists of tossing a coin 3 tintes and noting its outcome. A boy wins if all tosses give tite same outcome and loses otherwise. Find the probability that the boy loses the gante.
Solution:
If a coin is tossed 3 times then the total number of sample points n(S) = 2³ = 8
For winning the game all tosses gives the same outcome.
Let E be such that event then E = {HHH, TTT}
∴ n(E) = 2
P(E) = \(\frac{\mathrm{n}(\mathrm{E})}{\mathrm{n}(\mathrm{S})}=\frac{2}{8}\)
∴ The probability for the boy to loose the game
P\((\overline{\mathrm{E}})\) = 1 – P(E)
= 1 – \(\frac{2}{8}\) = \(\frac{6}{8}\).

Question 12.
If E₁, E₂ are two events with E₁∩ E₂ = Φ, then show that
\(P\left(E_1^C \cap E_2^C\right)=P\left(E_1^C\right)-P\left(E_2\right)\).
Solution:
\(P\left(E_1^C \cap E_2^C\right)=P\left(E_1^C\right)-P\left(E_2\right)\)
= 1 – [P(E₁ ∪ E₂)]
= 1 – [P(E₁) + P(E₂)]
[∵ P(E₁ ∩ E₂) = P(Φ) = 0]
(∵ From Addition theorem)
= 1 – P(E₁) – P(E₂)
= \(\mathrm{P}\left(\mathrm{E}_1^{\mathrm{C}}\right)\) – P(E₂).

II.
Question 1.
A pair of dice is rolled 24 times. A person wins by not getting a pair of 6’s on any ot the 24 rolls. What is the probability of his winnIng ? (This is the problem proved by the French gambler Chevalier de Mere (1607 – 1685) to Blaise Pascal, who in turn discussed it with Pierre de Fermat and solved the saine).
Solution:
When a pair of dice are rolled at a time, we get n(S) = 36.
If \(\overline{\mathrm{E}}\) is the event of getting a pair of 6’s then
n(\(\overline{\mathrm{E}}\)) = 1
∴ P(\(\overline{\mathrm{E}}\)) = \(\frac{1}{36}\)
∴ Probability of getting a pair of 6’s when two dice are roiled at a time = \(\frac{1}{36}\)
∴ Probability of not getting a pair of 6’s when two dice are rolled = P (E)
= 1 – P (\(\overline{\mathrm{E}}\))
= 1 – \(\frac{1}{36}\) = \(\frac{35}{36}\)
∴ A person wins the game if the dice are rolled for 24 times without getting pair of sixes.
∴ The probability that the person wins the game = \(\left(\frac{35}{36}\right)^{24}\).

Question 2.
If P is a probability function, then show that for any two events A and B.
P(A ∩ B) ≤ P(A) ≤ P(A ∪ B) ≤ P(A) + P(B)
Solution:
Here P(A ∩ B) = P(A) . P\(\left(\frac{\mathrm{B}}{\mathrm{A}}\right)\) where
P\(\left(\frac{\mathrm{B}}{\mathrm{A}}\right)\) is the conditional probability of B relative to A.
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
∴ P(A ∩ B) ≤ P(A) …………(i)
(∵ P\(\left(\frac{\mathrm{B}}{\mathrm{A}}\right)\) ≤ 1)
P(A) ≤ P(A ∪ B) …………..(ii)
(∵ A ⊆ A ∪ B)
∴ P(A ∪ B) ≤ P(A) + P(B) ………….(iii)
From (i), (ii) and (iii).
P(A ∩ B) ≤ P(A) ≤ P(A ∪ B) ≤ P(A) + P(B).

Question 3.
In a box contaIning 15 bulbs, 5 are defective. If 5 bulbs are selected at random from the box, find the probabflky of the event, that
i) none of them is defective.
ii) only one of theni is defective.
iii) atleast omie of ihietim is defective.
Solution:
Total number of bulbs = 15
Defective bulbs = 5
Number of good bulbs = 10
n(S) = ¹⁵C₅ = 3003.

i) None of theni is defective:
If E is the event of selecting all good bulbs from 10,
= ¹⁰C₂ = 252
∴ P(E) = \(\frac{n(E)}{n(S)}=\frac{252}{3003}=\frac{12}{143}\).

ii) Only one of them is defective:
So selecting one defective from 5 and 4 good bulbs from 10.
n(E₁) = ¹⁰C₄ . ⁵C₁ = 1050
∴ P(E₁) = \(\frac{\mathrm{n}\left(\mathrm{E}_1\right)}{\mathrm{n}(\mathrm{S})}=\frac{1050}{3003}=\frac{50}{143}\)

iii) Atleast one of them is defective:
P(E) = 1 – \(\frac{12}{143}\)
= \(\frac{131}{143}\).

Question 4.
A and B are seeldng admission into IIT. If the probability for A to be selected is 0.5 and that both to be selected is 0.3, then it is possible that, the probability of B to be selected is 0.9?
Solution:
Given that P(A) = 0.5; P (A ∩ B) = 0.3
and P(A ∪ B) ≤ 1
⇒ P (A) + P(B ) – P (A ∩ B) ≤ 1
⇒ 0.5 + P(B) – 0.3 ≤ 1
⇒ P( B) + 0.2 ≤ 1
⇒ P(B) ≤ 1 – 0.2 = 0.8
∴P (B) = 0.9 is not possible.

Question 5.
The probability for a contractor to get a road contract is \(\frac{2}{3}\) and to get a building contract is \(\frac{5}{9}\). The probability to get atleast one contract is \(\frac{4}{5}\). Find the probability that he gets both the contracts.
Solution:
Let A be the event that a contractor gets road contract and R he the event that a contractor gets a building contract.
Given P(A) = \(\frac{2}{3}\);
P(B) = \(\frac{5}{9}\)
P(A ∪ B) = \(\frac{4}{5}\)
We have to find P(A ∩ B).
From Addition theorem on probabilities, we have the probability that the contractor gets both the contracts is
P(A ∩ B) = P(A) + P(B) – P(A ∪ B)
= \(\frac{2}{3}+\frac{5}{9}-\frac{4}{5}=\frac{19}{45}\).

Question 6.
In a committee of 25 members, each member is proficient either in Mathematics or in Statistics or in both. If 19 of these are proficient in Mathematics, 16 in Statistics, find the probability that a person selected from the committee is proficient in both.
Solution:
Let A be the event that a person is proficient in Mathematics and B be the event that a person is proficient in Statistics.
n(S) = 25;
n(A) = 19;
n(B) = 16
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
P(A ∩ B) = P(A) + P(B) – P(A ∪ B)
= \(\frac{19}{25}+\frac{16}{25}\) – 1
= \(\frac{35-25}{25}=\frac{10}{25}=\frac{2}{5}\)
10 members arc proficient in Mathematics and Statistics.

Question 7.
A, B, C are three horses in a race. The probability of A to win the race is twice that of B and probability of B is twice that of C. What are the probablilties of A, B and C to win the race?
Solution:
Let probability of C to win the race be ‘x’
i.e.. P(C) = x
Given that P(B) = 2P(C) = 2x
and P(A) = 2P(B) = 2 × 2x = 4x
Sum of the probabilities is ‘1’.
x + 2x + 4x = 1
⇒ 7x = 1
⇒ x = \(\frac{1}{7}\)
∴ P(A) = 2(\(\frac{1}{7}\)) = \(\frac{2}{7}\)
and P(C) = x = \(\frac{1}{7}\).

Question 8.
A bag contains 12 two rupee coins, 7 one rupee coins and 4 half rupee coins. 1f three coins are selected at random, then find the probability that
i) the sum of three coins is maximum
ii) the sum of three is minimum
iii) each coin is of different value.
Solution:
Total number of coins 12 + 7 + 4 = 23 coins
n(S) = ²³C₃
i) The sum of the three coins is maximum when we select 3 two rupee coins from 12 coins in ¹²C₃ ways.
∴ n(E) = ¹²C₃
Probability that te sum of three coins is maximum.
P(E) = \(\frac{\mathrm{n}(\mathrm{E})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{12} C_3}{{ }^{23} C_3}\)

ii) The sum of the three coins is minimum when we selecting 3 half rupee coins from 4 half a rupee coins which can be done in ⁴C₃ ways.
∴ n(E) = ⁴C₃
∴ Probability for the sum of three coins to be minimum P(E) = \(\frac{{ }^4 C_3}{{ }^{23} C_3}\)

iii) Three coins each of different value
n(E) = ¹²C₁ . ⁷C₁ . ⁴C₁ and probability that the each coin has of different value = \(\frac{{ }^{12} C_1 \cdot{ }^7 C_1 \cdot{ }^4 C_1}{{ }^{23} C_3}\) ways.

Question 9.
The probabilities of three events A, B, C are such that P (A) = 0.3, P(B) = 0.4, P(C) = 0.8, P(A ∩ B) = 0.08, P(A ∩ C) = 0.28, P(A ∩ B ∩ C) = 0.09 and P(A ∩ B ∩ C) ≤ 0.75. Show that P(B ∩ C) lies in the interval [0.23, 0.48].
Solution:
Probability for any event is always less than or equal to ‘1’ and given that
P (A ∪ B ∪ C) ≥ 0.75
∴ 0.75 ≤ P(A ∪ B ∪ C) ≤ 1
⇒ 0.75 ≤ P(A) + P(B) + P(C) – P(A ∩ B) – P(B ∩ C) – P(C ∩ A) + P(A ∩ B ∩ C) ≤ 1
⇒ 0.75 ≤ 0.3 + 0.4 + 0.8 – 0.08 – P(B ∩ C) – 0.28 + 0.09 ≤ 1
⇒ 0.75 ≤ 1.23 – P(B ∩ C) ≤ 1
⇒ – 1.23 + 0.75 ≤ – P(B ∩ C) ≤ 1 – 1.23
⇒ – 0.48 ≤ – P(B ∩ C) ≤ – 0.23
⇒ 0.23 ≤ P(B ∩ C) ≤ 0.48
⇒ P(B ∩ C) ∈ [0.23, 0.48].

Question 10.
The probabilities of three mutually exclusive events are respeclively given as \(\frac{1+3 p}{3}, \frac{1-\mathbf{p}}{4}, \frac{1-2 \mathbf{p}}{2}\). Prove that \(\frac{1}{3} \leq p \leq \frac{1}{2}\).
Solution:
Suppose A, B, C are exclusive events
such that P(A) = \(\frac{1+3 \mathrm{P}}{3}\)
P(B) = \(\frac{1-P}{4}\)
P(C) = \(\frac{1-2 P}{2}\)
We know that
0 ≤ P(A) ≤ 1
o ≤ \(\frac{1+3 \mathrm{P}}{3}\) ≤ 1
0 ≤ 1 + 3P ≤ 3
– 1 ≤ 3P ≤ 3 – 1
\(\frac{-1}{3} \leq P \leq \frac{2}{3}\) ……………(1)
0 ≤ P(B) ≤ 1
0 ≤ \(\frac{1-\mathrm{P}}{4}\) ≤ 4
0 ≤ 1 – P ≤ 4
– 1 ≤ – P ≤ 4 – 1
1 ≥ P ≥ – 3
– 3 ≤ P ≤ 1 …………(2)
0 ≤ P(C) ≤ 1
0 ≤ \(\frac{1-2 \mathrm{P}}{2}\) ≤ 1
0 ≤ 1 – 2P ≤ 2
– 1 ≤ – 2P ≤ 2 – 1
1 ≥ 2P ≥ – 1
\(\frac{1}{2} \geq \mathrm{P} \geq-\frac{1}{2}\)
\(\frac{-1}{2} \leq P \leq \frac{1}{2}\) ……….(3)
Since A, B, C are exclusive events,
0 ≤ P(A ∪ B ∪ C) ≤ 1
⇒ 0 ≤ P(A) + P(B) + P(C) ≤ 1
⇒ 0 ≤ \(\frac{4+12 P+3-3 P+6-12 P}{12}\) ≤ 1
⇒ 0 ≤ \(\frac{13-3 P}{12}\) ≤ 1
⇒ 0 ≤ 13 – 3p ≤ 12
⇒ – 13 ≤ – 3P ≤ 12 – 13
⇒ 13 ≥ 3P ≥ 1
⇒ \(\frac{13}{3} \geq \mathrm{P} \geq \frac{1}{3}\)
⇒ \(\frac{1}{3} \leq \mathrm{P} \leq \frac{13}{3}\) ……………(4)
Max.of \(\left\{\frac{-1}{3},-3, \frac{-1}{2}, \frac{1}{3}\right\}=\frac{1}{3}\)
Min.of \(\left\{\frac{2}{3}, 1, \frac{1}{2}, \frac{13}{3}\right\}=\frac{1}{2}\)
(1), (2), (3) and (4) holds if \(\frac{1}{3} \leq P \leq \frac{1}{2}\).

Question 11.
On a festival day, a man plans to visit 4 holy temples A, B, C, D in a random order. Find the probability that he visits
(i) A before B
(ii) A before B and B before C
Solution:
Given that 4 holy temples are A, B, C and D.
Number of ways to visit 4 holy tempies in ⁴P₄ ways.
∴ n(S) = ⁴P₄
= 4! = 24

i) A before B:

ii) A before B and B before C:

Question 12.
From the employees of a company, 5 persons are selected to represent them in the managing committee  lite company. The particulars of 5 persons are as follows :

Name	Sex	Age in years
1. Harish	M	30
2. Rohan	M	33
3. Sheetala	F	46
4. Alis	F	28
5. Salim	M	41

 

A person is selected at random front this group to act as a spokesperson. Find the probability that the spokesperson will be either male or above 35 years.
Solution:
Let ‘A’ be the event of selecting a male
n(A) = 3
Let ‘B’ be the event of selecting a person whose age is above 35.
n(B) = 2
n(S) = 5, n(A ∩ B) = 1
P(A ∩ B) = P(A) + P(B) – P(A ∩ B)
= \(\frac{n(A)}{n(S)}+\frac{n(B)}{n(S)}-\frac{n(A \cap B)}{n(S)}\)
= \(\frac{3}{5}+\frac{2}{5}-\frac{1}{5}=\frac{4}{5}\).

Question 13.
Out of 100 Students, two sections of 40 and 60 are formed. If you and your friend are among the 100 students, find the probability that
(ï) you both enter the same section
(ii) you both enter the different sections.
Solution:
n(S) = ¹⁰⁰C₄₀
i) You both enter (lie saine seclion:
n(A) = ⁹⁸C₃₈ + ⁹⁸C₅₈
P(A) = \(\frac{n(A)}{n(S)}\)
= \(\frac{{ }^{98} C_{38}+{ }^{98} C_{58}}{{ }^{100} C_{40}}=\frac{17}{33}\)

ii) You both enter the different sections :
n(A) = ⁹⁸C₃₉ + ⁹⁸C₅₉
P(A) = \(\frac{n(A)}{n(S)}\)
= \(\frac{{ }^{98} C_{39}+{ }^{98} C_{59}}{{ }^{100} C_{40}}=\frac{16}{33}\).