## Book: Mathematics Part-II

### Chapter: 13. Probability

#### Subject: Maths - Class 12th

##### Q. No. 8 of Exercise 13.4

Listen NCERT Audio Books - Kitabein Ab Bolengi

8
##### A random variable X has the following probability distribution:X01234567P(X)0k2k2k3kK22 K27 K2 + kDetermine(i) k (ii) P(X < 3)(iii) P(X > 6) (iv) P(0 < X < 3)

Given: A random variable X with its probability distribution.

(i) As we know the sum of all the probabilities in a probability distribution of a random variable must be one. Hence the sum of probabilities of given table:

0 + k + 2k + 2k + 3k + k2 + 2k2 + 7K2 + k = 1

10K2 + 9k = 1

10K2 + 9k – 1 = 0

(10K-1)(k + 1) = 0 It is known that probability of any observation must always be positive that it can’t be negative.

So (ii) P(X < 3) = ?

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)

= 0 + k + 2k

= 3k (iii) P(X > 6) = ?

P(X > 6) = P(X = 7)

= 7K2 + k   (iv) P(0 < X < 3) = ?

P(0 < X < 3) = P(X = 1) + P(X = 2)

= k + 2k

= 3k 8

1
1
1
1
2
3
4
4
4
5
6
7
8
9
10
11
12
13
14
15
16
17