Math, asked by camjayasudha, 10 months ago


X is a binomial variable such that 2 P(X = 2) = P(X = 3) and mean of X is known to be
10/3. What would be the probability that X assumes at most the value 2?
(b) 17/81.
(c) 47/243
(d) 46/243.
(a) 16/81.​

Answers

Answered by amitnrw
20

Given : X is a binomial variable such that 2 P(X = 2) = P(X = 3) and mean of X is known to be 10/3

To find : What would be the probability that X assumes at most the value 2

Solution:

mean = 10/3

Mean = np  = 10/3

P(X) = ⁿCₓpˣ(1-p)ⁿ⁻ˣ

=> P(2)  =   ⁿC₂p²(1-p)ⁿ⁻²

   P(3) =      ⁿC₃p³(1-p)ⁿ⁻³

P(3) = 2 P(2)

=>  ⁿC₃p³(1-p)ⁿ⁻³  = 2 ⁿC₂p²(1-p)ⁿ⁻²

=> p/3!(n-3)!  = (2 /2!(n-2)! )(1 - p)

=> p/6   = (1-p)/(n - 2)

=> np  - 2p  = 6 - 6p

=> 4p = 6 - np

=> 4p = 6 - 10/3

=> 4p = 8/3

=> p  = 2/3

np = 10/3  => n = 5

probability that X assumes at most the value 2 = P(0) + P(1) + P(2)

 =   ⁵C₀(2/3)⁰(1/3)⁵ +   ⁵C₁(2/3)¹(1/3)⁴ +   ⁵C₂(2/3)²(1/3)³

=  1/243  +  10/243 +    40/243

= 51/243

= 17/81

17/81  is the  probability that X assumes at most the value 2

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