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Standard XII

Mathematics

Classical Definition of Probability

A random vari...

Question

A random variable X has the following probability distribution.

X

0

1

2

3

4

5

6

7

P (X)

0

k

2k

2k

3k

k²

2k²

7k² + k

Determine

(i) k

(ii) P (X < 3)

(iii) P (X > 6)

(iv) P (0 < X < 3)

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Solution

(i) It is known that the sum of probabilities of a probability distribution of random variables is one.

k = − 1 is not possible as the probability of an event is never negative.

∴

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

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

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

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A random variable X has the following probability distribution. X 0 1 2 3 4 5 6 7 P (X) 0 k 2k 2k 3k k2 2k2 7k2 + k Determine (i) k (ii) P (X < 3) (iii) P (X > 6) (iv) P (0 < X < 3)

A random variable X has the following probability distribution.

X

0

1

2

3

4

5

6

7

P (X)

0

k

2k

2k

3k

k²

2k²

7k² + k

Determine

(i) k

(ii) P (X < 3)

(iii) P (X > 6)

(iv) P (0 < X < 3)