RD Sharma Solutions for Class 12 Maths Chapter 31 Probability Exercise 31.7 are given here for students who aspire to kick-start preparations for the board examination. These solutions contain shortcut tips and tricks to help students achieve high marks in the exam. Further, students are recommended to practise the RD Sharma Class 12 Solutions regularly to attain a strong grip on solving problems. Also, they can now download these solutions in PDF format for free from the link provided below.
RD Sharma Solutions for Class 12 Maths Chapter 31 Exercise 7
Access RD Sharma Solutions for Class 12 Maths Chapter 31 Exercise 7
Exercise 31.7
1. Solution:
Given,
Urn I contains 1 white, 2 black and 3 red balls.
Urn II contains 2 white, 1 black and 1 red balls.
Urn III contains 4 white, 5 black and 4 red balls.
Let E1, E2, E3 and A be the events as defined:
E1 = Selecting urn I
E2 = Selecting urn II
E3 = Selecting urn III
A = Drawing 1 white ball and 1 red ball
Now.
P(E1) = P(E2) = P(E3) = 1/3 [As there are only 3 urns]
And,
P(A/E1) = P(Drawing 1 red ball and 1 white ball from urn I)
= (1C1 x 3C1)/ 6C2
= (1 x 3)/ (6 x 5/2)
= 1/5
P(A/E2) = P(Drawing 1 red ball and 1 white ball from urn II)
= (2C1 x 1C1)/ 4C2
= (2 x 1)/ (4 x 3/2)
= 1/3
P(A/E3) = P(Drawing 1 red ball and 1 white ball from urn III)
= (4C1 x 3C1)/ 12C2
= (4 x 3)/ (12 x 11)/2
= 2/11
We have to find,
P(both balls came from urn I) = P(E1/A)
P(both balls came from urn II) = P(E2/A)
P(both balls came from urn III) = P(E3/A)
So, by Baye’s theorem, we get
And,
Therefore, the required probabilities are 33/118, 55/118 and 30/118.
2. Solution:
Given,
Bag A contains 2 white and 3 red balls.
Bag B contains 4 white and 5 red balls.
Let E1, E2 and A be events as given below.
E1 – Choosing bag A
E2 – Choosing bag B
A – Drawing one red ball
So, we have
P(E1) = P(E2) = 1/2 [Since, there are only 2 bags]
Now,
P(A/E1) = P(drawing a red ball from Bag A)
= 3/5
P(A/E2) = P(drawing a red ball from Bag B)
= 5/9
So, by Baye’s theorem, we get
Hence, the required probability is 25/52.
3. Solution:
Given,
Urn I contains 2 white and 3 black balls.
Urn II contains 3 white and 2 black balls.
Urn III contains 4 white and 1 black balls.
Let E1, E2, E3 and A be the events as defined:
E1 = Selecting urn I
E2 = Selecting urn II
E3 = Selecting urn III
A = Drawing 1 white ball
Now.
P(E1) = P(E2) = P(E3) = 1/3 [As there are only 3 urns]
And,
P(A/E1) = P(Drawing one white ball from urn I)
= 2/5
P(A/E2) = P(Drawing one white ball from urn II)
= 3/5
P(A/E3) = P(Drawing one white ball from urn III)
= 4/5
According to the question, we need to find
P(Drawn one white ball is from urn 1) = P(E1/A)
So, By Baye’s theorem, we get
Therefore, the required probability is 2/9.
4. Solution:
Given,
Urn I contains 7 white and 3 black balls.
Urn II contains 4 white and 6 black balls.
Urn III contains 2 white and 8 black balls.
Let E1, E2, E3 and A be the events as defined:
E1 = Selecting urn I
E2 = Selecting urn II
E3 = Selecting urn III
A = Drawing 2 white balls without replacement
Also given,
P(E1) = 0.20
P(E2) = 0.60
P(E3) = 0.20
Now,
P(A/E1) = P(Drawing two white balls from urn I)
= 7C2/10C2
= (7×6)/2 ÷ (10×9)/2
= 7/15
P(A/E2) = P(Drawing two white balls from urn II)
= 4C2/10C2
= (4×3)/2 ÷ (10×9)/2
= 12/90
= 2/15
P(A/E3) = P(Drawing two white balls from urn III)
= 2C2/10C2
= 1 ÷ (10×9)/2
= 1/45
According to the question, we need to find
P(Drawn two white balls are from urn III) = P(E3/A)
So, By Baye’s theorem, we get
Thus, the required probability is 1/40.
5. Solution:
Let’s consider the following events,
E1 – Getting 1 or 2 in a throw of die.
E2 – Getting 3, 4, 5 or 6 in a throw of die.
A – Getting exactly one tail
So clearly, we have
P(E1) = 2/6 = 1/3
P(E2) = 4/6 = 2/3
P(A/E1) = 3/8
P(A/E2) = ½
Now, the required probability is given by
6. Solution:
Let’s consider the following events:
E1 – The first group wins
E2 – The second group wins
A – New product is introduced
It’s given that,
P(E1) = 0.6
P(E2) = 0.4
P(A/E1) = 0.7
P(A/E2) = 0.3
So, the required probability P(E2/A) is given by
Hence,
P(E2/A) = 2/9
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