Chapter 13 Probability Ex 13.4 Solutions
Question - 11 : - Two dice are thrown simultaneously. If X denotes the number of sixes, find the expectation of X.
Answer - 11 : -
Given a die is throwntwo times.
When a die is tossedtwo times then the number of observations will be (6 × 6) = 36.
Now, let X is a randomvariable which represents the success and is given as six appears on at leastone die.
Now
P (X = 0) = P (sixdoes not appear on any of die) = 5/6 × 5/6 = 25/36
P (X = 1) = P (sixappears at least once of the die) = (1/6 × 5/6) + (5/6 × 1/6) = 10/36 = 5/18
P (X = 2) = P (sixdoes appear on both of die) = 1/6 × 1/6 = 1/36
Hence, the requiredprobability distribution is,
| X | 0 | 1 | 2 |
| P (X) | 25/36 | 5/18 | 1/36 |
Therefore Expectationof X E (X):
Question - 12 : - Two numbers are selected at random (without replacement) from the first six positive integers. Let X denote the larger of the two numbers obtained. Find E(X).
Answer - 12 : -
Given first sixpositive integers.
Two numbers can beselected at random (without replacement) from the first six positive integer in6 × 5 = 30 ways.
X denote the larger ofthe two numbers obtained. Hence, X can take any value of 2, 3, 4, 5 or 6.
For X = 2, thepossible observations are (1, 2) and (2, 1)
P (X) = 2/30 = 1/15
For X = 3, thepossible observations are (1, 3), (3, 1), (2, 3) and (3, 2).
P (X) = 4/30 = 2/15
For X = 4, thepossible observations are (1, 4), (4, 1), (2,4), (4,2), (3,4) and (4,3).
P (X) = 6/30 = 1/5
For X = 5, thepossible observations are (1, 5), (5, 1), (2,5), (5,2), (3,5), (5,3) (5, 4) and(4,5).
P (X) = 8/30 = 4/15
For X = 6, thepossible observations are (1, 6), (6, 1), (2,6), (6,2), (3,6), (6,3) (6, 4),(4,6), (5,6) and (6,5).
P (X) = 10/30 = 1/3
Hence, the requiredprobability distribution is,
| X | 2 | 3 | 4 | 5 | 6 |
| P (X) | 1/15 | 2/15 | 1/5 | 4/15 | 1/3 |
Therefore E(X) is:
Question - 13 : - Let X denote the sum of the numbers obtained when two fair dice are rolled. Find the variance and standard deviation of X.
Answer - 13 : -
Given two fair diceare rolled
When two fair dice arerolled then number of observations will be 6 × 6 = 36.
X denote the sum ofthe numbers obtained when two fair dice are rolled. Hence, X can take any valueof 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 or 12.
For X = 2, thepossible observations are (1, 1).
P (X) = 1/36
For X = 3, thepossible observations are (1, 2) and (2, 1)
P (X) = 2/36 = 1/18
For X = 4, thepossible observations are (1, 3), (2, 2) and (3, 1).
P (X) = 3/36 = 1/12
For X = 5, thepossible observations are (1, 4), (4, 1), (2, 3) and (3, 2)
P (X) = 4/39 = 1/9
For X = 6, thepossible observations are (1, 5), (5, 1), (2, 4), (4, 2) and (3, 3).
P (X) = 5/36
For X = 7, thepossible observations are (1, 6), (6, 1), (2,5), (5,2),(3,4) and (4,3).
P (X) = 6/36 = 1/6
For X = 8, thepossible observations are (2, 6), (6, 2), (3, 5), (5, 3) and (4, 4).
P (X) = 5/36
For X = 9, thepossible observations are (5, 4), (4, 5), (3, 6) and (6, 3)
P (X) = 4/36 = 1/9
For X = 10, thepossible observations are (5, 5), (4, 6) and (6, 4).
P (X) = 3/36 = 1/12
For X = 11, thepossible observations are (6, 5) and (5, 6)
P (X) = 2/36 = 1/18
For X = 12, thepossible observations are (6, 6).
P (X) = 1/36
Hence, the requiredprobability distribution is,
| X | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
| P (X) | 1/36 | 1/18 | 1/12 | 1/9 | 5/36 | 1/6 | 5/36 | 1/9 | 1/12 | 1/18 | 1/36 |
Therefore E(X) is:
= √5.833
= 2.415
Question - 14 : - A class has 15 students whose ages are 14, 17, 15, 14, 21, 17, 19, 20, 16, 18, 20, 17, 16, 19 and 20 years. One student is selected in such a manner that each has the same chance of being chosen and the age X of the selected student is recorded. What is the probability distribution of the random variable X? Find mean, variance and standard deviation of X.
Answer - 14 : -
Given the class of 15students with their ages.
Form the given data wecan draw a table
| X | 14 | 15 | 16 | 17 | 18 | 19 | 20 | 21 |
| f | 2 | 1 | 2 | 3 | 1 | 2 | 3 | 1 |
P(X = 14) = 2/15
P(X = 15) = 1/15
P(X = 16) = 2/15
P(X = 17) = 3/15
P(X = 18) = 1/15
P(X = 19) = 2/15
P(X = 20) = 3/15
P(X = 21) = 1/15
Hence, the requiredprobability distribution is,
| X | 14 | 15 | 16 | 17 | 18 | 19 | 20 | 21 |
| P (X) | 2/15 | 1/15 | 2/15 | 3/15 | 1/15 | 2/15 | 3/15 | 1/15 |
Therefore E(X) is:
Question - 15 : - 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 E(X) and Var (X).
Answer - 15 : -
Given: X = 0 ifmembers oppose, and X = 1 if members are in favour.
P(X = 0) = 30% =30/100 = 0.3
P(X = 1) = 70% =70/100 = 0.7
Hence, the requiredprobability distribution is,
Therefore E(X) is:
= 0 × 0.3 + 1 × 0.7
⇒ E(X) = 0.7
And E(X2)is:
= (0)2 ×0.3 + (1)2 × 0.7
⇒ E(X2)= 0.7
Then Variance, Var(X)= E(X2) – (E(X))2
= 0.7 – (0.7)2
= 0.7 – 0.49 = 0.21
Question - 16 : - The mean of the numbers obtained on throwing a die having written 1 on three faces, 2 on two faces and 5 on one face is
A. 1
B. 2
C. 5
D. 8/3
Answer - 16 : -
B. 2
Explanation:
Given a die havingwritten 1 on three faces, 2 on two faces and 5 on one face.
Let X be the randomvariable representing a number on given die.
Then X can take anyvalue of 1, 2 or 5.
The total numbers issix.
Now
P(X = 1) = 3/6 = ½
P(X = 2) = 1/3
P(X = 5) = 1/6
Hence, the requiredprobability distribution is,
| X | 1 | 2 | 5 |
| P (X) | 1/2 | 1/3 | 1/6 |
Therefore Expectationof X E(X):
Question - 17 : - Suppose that two cards are drawn at random from a deck of cards. Let X be the number of aces obtained. Then the value of E(X) is
A. 37/221
B. 5/13
C. 1/13
D. 2/13
Answer - 17 : -
D. 2/13
Explanation:
Given a deck of cards.
X be the number of aces obtained.
Hence, X can take value of 0, 1 or 2.
As we know, in a deck of 52 cards, 4 cards are aces. Therefore 48 cards are non- ace cards.
= 6/1326
Hence, the requiredprobability distribution is,
| X | 0 | 1 | 2 |
| P (X) | 1128/1326 | 192/1326 | 6/1326 |
Therefore Expectationof X E(X):
