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Question 27 of 43

Q.The probability distribution of a random variable XX is as follows: p(x)=C(x2+x),  x=−2,1,2p(x) = C(x^2 + x), \; x = -2, 1, 2. Find the value of CC.

(OR)
State the properties of Bernoulli trials. (Any two)
Gujarat GsebGujarat Board (GSEB) HSC Commerce Board 2023Subjective· 2mImportance★★★★★
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∑p(x)=2C+2C+6C=10C=1⇒C=0.1\sum p(x)=2C+2C+6C=10C=1\Rightarrow C=0.1. OR: Bernoulli trials have two outcomes, are independent, with constant pp and a fixed number of trials.

Main part: p(x)=C(x2+x)p(x)=C(x^2+x) for x=−2,1,2x=-2,1,2. Evaluate each:

p(−2)=C((−2)2+(−2))=C(4−2)=2C,p(-2)=C\big((-2)^2+(-2)\big)=C(4-2)=2C,

p(1)=C(1+1)=2C,p(2)=C(4+2)=6C.p(1)=C(1+1)=2C,\qquad p(2)=C(4+2)=6C.

Since the probabilities must sum to 11:

2C+2C+6C=10C=1  ⇒  C=110=0.1.2C+2C+6C=10C=1\;\Rightarrow\;C=\frac{1}{10}=0.1.

OR part — properties of Bernoulli trials (any two):

  1. Each trial has exactly two mutually exclusive outcomes — success and failure.
  2. The trials are independent of one another. …

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