Inter 1st Year Maths Probability Solutions Exercise 14c

Practicing the AP Board Solutions Class 11 Maths and Chapter 14 Inter 1st Year Maths Probability Solutions Exercise 14c Pdf Download will help students to clear their doubts quickly.

Intermediate 1st Year Maths Probability Solutions Exercise 14c

Probability Exercise 14c Solutions

Probability Class 11 Exercise 14c Solutions – Probability 14c Exercise Solutions

I.

Question 1.
A box contains 10 red marbles, 20 blue marbles, and 30 green marbles. 5 marbles are drawn at random from the box. What is the probability that
(i) All will be blue?
(ii) At atleast one will be green?
Solution:
Total number of marbles = 10 + 20 + 30 = 60
Number of ways of drawing 5 marbles from 60 marbles = ⁶⁰C₅
(i) All the drawn marbles will be blue if we draw 5 marbles out of 20 blue marbles.
5 blue marbles can be drawn from 20 blue marbles in ²⁰C₅ ways.
∴ Probability that all marbles will be blue = \(\frac{{ }^{20} \mathrm{C}_5}{{ }^{60} \mathrm{C}_5}\)

(ii) Number of ways in which the drawn marble is not green = ^(20+10)C₅ = ³⁰C₅
∴ Probability that no marble is green = \(\frac{{ }^{30} C_5}{{ }^{60} C_5}\)
∴ Probability that atleast one marble is green = 1 – \(\frac{{ }^{30} C_5}{{ }^{60} C_5}\)

Question 2.
4 cards are drawn at random from a well-shuffled deck of 52 cards. What is the probability of obtaining 3 diamonds and one spade?
Solution:
Number of ways of drawing 4 cards from 52 cards = ⁵²C₄
In a deck of 52 cards, there are 13 diamonds and 13 spades.
∴ Number of ways of drawing 3 diamonds and one spade = ¹³C₃ × ¹³C₁
Thus, the probability of obtaining 3 diamonds and one spade = \(\frac{{ }^{13} \mathrm{C}_3 \times{ }^{13} \mathrm{C}_1}{{ }^{52} \mathrm{C}_4}\)

Question 3.
A die has two faces each with the number ‘1’, three faces each with the number ‘2’, and one face with the number ‘3’. If the die is rolled once, determine
(i) P(2)
(ii) P(1 or 3)
(iii) P(not 3)
Solution:
Total number of faces = 6
(i) Number faces with number 2 = 3
∴ P(2) = \(\frac{3}{6}=\frac{1}{2}\)

(ii) P(1 or 3) = P(not 2)
= 1 – P(2)
= 1 – \(\frac {1}{2}\)
= \(\frac {1}{2}\)

(iii) Number of faces with number 3 = 1
Thus, P(not 3) = 1 – P(3)
= 1 – \(\frac {1}{6}\)
= \(\frac {5}{6}\)

Question 4.
In a certain lottery, 10,000 tickets are sold, and ten equal prizes are awarded. What is the probability of not getting a prize if you buy
(i) one ticket
(ii) two tickets
(iii) 10 tickets
Solution:
Total number of tickets sold = 10,000
Number prizes awarded = 10
(i) If we buy one ticket, then
P(getting a prize) = \(\frac {10}{10000}\) = \(\frac {1}{1000}\)
∴ P(not getting a prize) = 1 – \(\frac {1}{1000}\) = \(\frac {999}{1000}\)

(ii) If we buy two tickets, then
Number of tickets not awarded = 10,000 – 10 = 9990.
∴ P(not getting a prize) = \(\frac{{ }^{9990} \mathrm{C}_2}{{ }^{10000} \mathrm{C}_2}\)

(iii) If we buy 10 tickets, then
∴ P(not getting a prize) = \(\frac{{ }^{9990} \mathrm{C}_{10}}{{ }^{1000} \mathrm{C}_{10}}\)

Question 5.
A and B are two events such that P(A) = 0.54, P(B) = 0.69, and P(A ∩ B) = 0.35. Find
(i) P(A ∪ B)
(ii) P(A’ ∩ B’)
(iii) P(A ∩ B’)
(iv) P(B ∩ A’)
Solution:
P(A) = 0.54, P(B) = 0.69, P(A ∩ B) = 0.35
(i) P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= 0.54 + 0.69 – 0.35
= 0.88

(ii) P(A^C ∩ B^C) = 1 – P(A ∪ B)
= 1 – 0.88
= 0.12

(iii) P(A ∩ B^C) = P(A) – P(A ∩ B)
= 0.54 – 0.35
= 0.19

(iv) P(A^C ∩ B) = P(B) – P(A ∩ B)
= 0.69 – 0.35
= 0.34

Question 6.
Three letters are dictated to three persons, and an envelope is addressed to each of them. The letters are inserted into the envelopes at random so that each envelope contains exactly one letter. Find the probability that at least one letter is in its proper envelope.
Solution:
3 letters can go into any of the 3 envelopes = 3! = 6
For 3 items, the number of derangements is

II.

Question 1.
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 section?
(b) You both enter the different sections?
Solution:
My friend and I are among the 100 students.
Total number of ways of selecting 2 students out of 100 students = ¹⁰⁰C₂
(a) The two of us will enter the same section if both of us are among 40 students or among 60 students.
∴ Number of ways in which both of us enter the same section = ⁴⁰C₂ + ⁶⁰C₂
∴ The probability that both of us enter the same section

(b) P(We enter different sections) = 1 – P(We enter the same section)
= 1 – \(\frac {17}{33}\)
= \(\frac {16}{33}\)

Question 2.
From the employees of a company, 5 persons are selected to represent them in the managing committee of the company. Particulars of five persons are as follows:

A person is selected at random from this group to act as a spokesperson. What is the probability that the spokesperson will be either male or over 35 years old?
Solution:
Let E be the event in which the spokesperson will be a male, and F be the event in which the spokesperson will be over 35 years of age.
Accordingly, P(E) = \(\frac {3}{5}\) and P(F) = \(\frac {2}{5}\)
Since there is only one male who is over 35 years of age,
P(E ∩ F) = \(\frac {1}{5}\)
P(E ∪ F) = P(E) + P(F) – P(E ∩ F)
= \(\frac{3}{5}+\frac{2}{5}-\frac{1}{5}\)
= \(\frac {4}{5}\)
Thus, the probability that the spokesperson will either be a male or over 35 years of age is \(\frac {4}{5}\).

Question 3.
If 4-digit numbers greater than 5,000 are randomly formed from the digits 0, 1, 3, 5, and 7, what is the probability of forming a number divisible by 5 when
(i) The digits are repeated?
(ii) The repetition of digits is not allowed.
Solution:
Forming 4-digit numbers greater than 5000 using the digits {0, 1, 3, 5, 7}
A number divisible by 5 must end in 0 or 5.
First digit (thousand place) must be ≥ 5.
Only digits 5 or 7 can be used, 2 choices
Other 3 digits (hundreds, tens, units) = 5 × 5 × 5 = 125 choices
Total numbers (with repetition) = 2 × 125 = 250.

Last digit is 0:
Units digits = 0 ⇒ 1 way
A thousand digit must be 5 or 7 = 2 choices
Hundreds and terms digits = 5 × 5 = 25
ending in 0 = 2 × 25 = 50

Last digit is 5:
Units digit = 5 ⇒ 1 way
Thousand digit = 5 or 7
Hundreds and ten digits = 5 × 5 = 25
ending in 5 = 2 × 25 = 50.
Total favourable numbers (ending 0 or 5) = 50 + 50 = 100
P = \(\frac{\text { Favourable }}{\text { Total }}=\frac{100}{250}=\frac{2}{5}\)
Total 4-digit numbers greater than 5000 (No repetition)
First digit (thousand place) must be 5 or 7 = 2 choices
Repeating 3 digits is ⁴P₃ = 4 × 3 × 2 = 24
Total (No repetition) = 2 × 24 = 48

Ending 0:
Units digit = 0
Thousands digit: must be 5 or 7
choose 2 from 3 than ³P₂ = 3 × 2 = 6
ending in 0 = 2 × 6 = 12

Ending in 5:
Units digit = 5
Thousand digit: must be 5 or 7
chose 2 from 3 then ³P₂ = 3 × 2 = 6
ending is 5 = 1 × 6 = 6
Total favourable numbers = 12 + 6 = 18
P = \(\frac{\text { Favourable }}{\text { Total }}=\frac{18}{48}=\frac{3}{8}\)

Question 4.
The number lock of a suitcase has 4 wheels, each labelled with ten digits, i.e., from 0 to 9. The lock opens with a sequence of four digits with no repeats. What is the probability of a person getting the right sequence to open the suitcase?
Solution:
The number lock has 4 wheels, each labelled with ten digits, i.e., from 0 to 9.
Number of ways of selecting 4 different digits out of the 10 digits: ¹⁰C₄.
Now, each combination of 4 different digits can be arranged in 4! ways.
∴ Number of four digits with no repetitions = ¹⁰C₄ × 4!
= \(\frac{10!}{4!6!}\) × 4!
= 7 × 8 × 9 × 10
= 5040
There is only one number that can open the suitcase.
Thus, the required probability is \(\frac {1}{5040}\).