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The probability distribution of a random variable X is given below: (i) Determine the value of k. (ii) Determine \[\mathbf{P}\left( \mathbf{X}\text{ }\text{£}\text{ }\mathbf{2} \right)\] and \[\mathbf{P}\left( \mathbf{X}\text{ }>\text{ }\mathbf{2} \right)\] (iii) Find \[\mathbf{P}\left( \mathbf{X}\text{ }\text{£}\text{ }\mathbf{2} \right)\text{ }+\text{ }\mathbf{P}\text{ }\left( \mathbf{X}\text{ }>\text{ }\mathbf{2} \right)\]
The probability distribution of a random variable X is given below: (i) Determine the value of k. (ii) Determine \[\mathbf{P}\left( \mathbf{X}\text{ }\text{£}\text{ }\mathbf{2} \right)\] and \[\mathbf{P}\left( \mathbf{X}\text{ }>\text{ }\mathbf{2} \right)\] (iii) Find \[\mathbf{P}\left( \mathbf{X}\text{ }\text{£}\text{ }\mathbf{2} \right)\text{ }+\text{ }\mathbf{P}\text{ }\left( \mathbf{X}\text{ }>\text{ }\mathbf{2} \right)\]
CBSE | Exercise 1 | Maths | NCERT Exemplar | Probability
The probability distribution of a random variable X is given below: (i) Determine the value of k. (ii) Determine \[\mathbf{P}\left( \mathbf{X}\text{ }\text{£}\text{ }\mathbf{2} \right)\] and \[\mathbf{P}\left( \mathbf{X}\text{ }>\text{ }\mathbf{2} \right)\] (iii) Find \[\mathbf{P}\left( \mathbf{X}\text{ }\text{£}\text{ }\mathbf{2} \right)\text{ }+\text{ }\mathbf{P}\text{ }\left( \mathbf{X}\text{ }>\text{ }\mathbf{2} \right)\]

(i) W.k.t \[P\left( 0 \right)\text{ }+\text{ }P\left( 1 \right)\text{ }+\text{ }P\left( 2 \right)\text{ }+\text{ }P\left( 3 \right)\text{ }=\text{ }1\]

\[\Rightarrow k\text{ }+\text{ }k/2\text{ }+\text{ }k/4\text{ }+\text{ }k/8\text{ }=\text{ }1\]

\[\left( 8k\text{ }+\text{ }4k\text{ }+\text{ }2k\text{ }+\text{ }k \right)/8\text{ }=\text{ }1\]

\[15k\text{ }=\text{ }8\]

Hence, \[k=8/15\]

(ii) \[P\left( X\text{ }\le \text{ }2 \right)\text{ }=\text{ }P\left( X\text{ }=\text{ }0 \right)\text{ }+\text{ }P\left( X\text{ }=\text{ }1 \right)\text{ }+\text{ }P\left( X\text{ }=\text{ }2 \right)\]

= k + k/2 + k/4 = 7k/4 = 7/4 x 8/15 = 14/15

\[=\text{ }k\text{ }+\text{ }k/2\text{ }+\text{ }k/4\text{ }=\text{ }7k/4\text{ }=\text{ }7/4\text{ }x\text{ }8/15\text{ }=\text{ }14/15\]

And \[P\left( X\text{ }>\text{ }2 \right)\text{ }=\text{ }P\left( X\text{ }=\text{ }3 \right)\text{ }=\text{ }k/8\text{ }=\text{ }1/8\text{ }x\text{ }8/15\text{ }=\text{ }1/15\]

(iii) \[P\left( X\text{ }\le \text{ }2 \right)\text{ }+\text{ }P\left( X\text{ }\ge \text{ }2 \right)\text{ }=\text{ }14/15\text{ }+\text{ }1/15\text{ }=\text{ }\left( 14\text{ }+\text{ }1 \right)/15\text{ }=\text{ }15/15\text{ }=\text{ }1\]

i

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