RBSE Solutions for Class 12 Maths Chapter 16 Probability and Probability Distribution Ex 16.4
RBSE Solutions for Class 12 Maths Chapter 16 Probability and Probability Distribution Ex 16.4
Rajasthan Board RBSE Class 12 Maths Chapter 16 Probability and Probability Distribution Ex 16.4
Question 1.
Find which of the following probability distribution is possible for a random variable :
Solution:
(i) Sum of probabilities
= 0.4 + 0.4 + 0.2 = 1
Hence, given distribution is a probability distribution.
(ii) Sum of probabilities = 0.6 + 0.1 + 0.2
= 0.9 ≠ 1
Hence, the given distribution is not a probability distribution.
(iii) Here, one of the probability is – 0.1. which is negative. Hence, this distribution is not probability distribution.
Question 2.
Find the probability distribution of X, the number of heads in a simultaneous toss of two coins.
Solution:
Possible values of X are 0, 1, or 2
Now P(X = 0) = P(No head)
= P(Tail first time and tail second time)
= P(Tail first time).P(Tail in second time)
= 1/2 × 1/2 = 1/4
= P(TH or HT) = P (TH) + P(HT)
= 1/4 + 1/4 = 1/2
P(x = 2) = P (2 Head) = P(HH) = 1/4
∴ Required distribution is
Question 3.
4 rotten oranges are mixed accidently with 16 good oranges. Find the probability distribution of the number of rotten oranges in a draw of two oranges.
Solution:
4 bad oranges are mixed with 16 good oranges The number of total oranges = 4 + 16 = 20
2 bad oranges are to be choosen
∴ Probability of choosing a bad orange
= 4/20 = 1/5
∴ Probability of choosing one good orange
= 1 – 1/5 = 4/5
∵ X is the number of bad oranges
Question 4.
An urn contains 4 white and 3 red bails. Find the probability distribution of the number of red balls if 3 balls are drawn at random.
Solution:
Three balls are taken out of an urn
∴ Sample space is
= S {RRR, RRW, RWR, WRR, RWW WRW, WWR WWW}
Where R respresents red ball and W, white ball.
Let X = number of red balls. So possible value of X are 3,2,1,2,1,0 or 0, 1,2, 3 (No red ball)
Question 5.
From a lot of 10 items containing 3 defectives, a sample of 4 items is drawn at random. Let the random variable X denote the number of defective items in the sample. If the sample is drawn randomly, find
(i) The probability distribution of X
(ii) P(x ≤ 1)
(iii) P(x < 1)
(iv) P(0 < x < 2)
Solution:
Given, from a lot of 10 items, 3 are defective
∴ Good items = 10 – 3 = 7
Let x represents the number of defective items.
Clearly values of X are 0, 1,2, 3.
P(x = 0) = P(GGGG)
= p (good items)
= 7/10 × 6/9 × 5/8 × 4/7 = 1/6
P (x = 1) = P (one good and three defective)
Question 6.
A die is manufactured in such a way that an even number is twice likely to occur as an odd number. If the die is tossed twice, find the probability distribution of the random variable X representing the perfect square in the two tosses.
Solution:
Let X represents the number of perfect squares.
S = sample space = { 1, 2, 3. 4, 5, 6}
∴ n(S) 6
n(X) = Possible number of perfect squares = {1,4} = 2
∴ Probability of getting perfect square
Question 7.
An urn contains 4 white and 6 red balls. Four balls are drawn at random from the urn. Find the probability of the number of white balls.
Solution:
Let X represents white balls.
∴ Total balls =4 + 6 = 10
Four balls are drawn at random.
∴ Expected values of X will be 0,1,2,3,4.
∴ P(X = 0) = P(all red)
= P(RRRR)
Question 8.
Find the probability distribution of number doublets in three throws of a pair of dice.
Solution:
Let x = number of doublets
∴ Expected value of x will be 0, 1, 2, 3
Set of all doublets available on a pair of dice in a throw = (1,1), (2, 2), (3,3), (4,4), (5, 5), (6, 6)
Number of all possible ways on throwing a pair of dice. = 6 × 6 = 36
∴ Probablity of getting a doublets on a pair of dice in a throw = 6/36 = 1/6
∴ Probablity of getting a doublets on a pair of dice throw = 1 – 1/6 = 5/6
Question 9.
Let a pair of dice be thrown and the random variable X be the sum of the numbers that appear on the two dice. Find the mean of X.
Solution:
The number of possible outcomes when two dice are thrown = 6 × 6 = 36
Question 10.
Find the variance of the number obtained on a throw of an unbiased die.
Solution:
Let sample space of observation be s = {1,2, 3, 4, 5, 6}
X represents the number appeared on dice. Then x is a random variable takes value 1, 2, 3,4, 5 or 6.
Also P(1) = P(2) = P(3) = P(4) = P(5) = P(6) = 1/6
∴ Probablity distribution is Following:
Question 11.
In a meeting, 70% of the members favour and 30% oppose a certain proposal. A member is selected at random and we take X = 0 if he opposed, and X = 1 if he is in favour. Find the mean and variance of X.
Solution:
Probability of members, favouring any proposal on X = 1
70% = 70/100 = 0.70
Probability of members, opposing any proposal on X= 0
30% = 30/100 = 0.30
∴ Required probability distribution is following
Question 12.
Two card are drawn simultaneously (or successively without replacement) from a well shuffled pack of 52 cards. Find the mean, variance and standard deviation of the number of kings.
Solution:
Two cards are drawn
∴ Number of ways that two cards are not king.
Number of ways of drawing two cards out of 52 cards = 52C2
∴ Probability of not drawing any king = 1128/1326
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