Test Index

Probability Part 3

Section: Mathematics

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Question : 17 of 35

Marks: +1, -0

If a random variable

x

has the probability distribution

$x$	0	1	2	3	4	5	6	7
$P(x)$	0	2 k	k	3 k	$2k^{2}$	2 k	$k^{2}+k$	$7k^{2}$

P(3 < x \le 6)

[22 Jan 2026 Shift 1]

0.34
0.64
0.22
0.33

Solution:

Concept:

The sum of all probabilities in a probability distribution is always 1.

Formula:

\sum P(x) = 1

Solution:

Write the sum of all probabilities:

2k + k + 3k + 2k^2 + 2k + (k^2 + k) + 7k^2 = 1

Combine like terms:

k

terms:

2k + k + 3k + 2k + k = 9k

k^2

terms:

2k^2 + k^2 + 7k^2 = 10k^2

So

10k^2 + 9k = 1

Rearrange:

10k^2 + 9k - 1 = 0

Factor:

(10k - 1)(k + 1) = 0

Thus

k = \frac{1}{10}

or

k = -1

(reject as probability cannot be negative).

Hence

k = 0.1

.

Now find

P(3 < x \le 6)

which is

P(x=4) + P(x=5) + P(x=6)

:

P(4) = 2k^2 = 2(0.01) = 0.02

P(5) = 2k = 2(0.1) = 0.2

P(6) = k^2 + k = 0.01 + 0.1 = 0.11

Sum =

0.02 + 0.2 + 0.11 = 0.33

Answer:

P(3 < x \le 6) = 0.33

, which corresponds to option D.

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