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Chapter 7: Q.7.10 (page 352)

Consider 3 trials, each having the same probability of success. Let Xdenote the total number of successes in these trials. If E[X]=1.8, what is

(a) the largest possible value of PX=3?

(b) the smallest possible value of P{X=3}}?

Short Answer

Expert verified
  1. The required largest possible value is P(X=3)=0.6.
  2. The lowest possible value isP(X=3)=0.

Step by step solution

01

Given information (Part a)

Consider 3trials.

X=total number of success

E[X]=1.8

02

Solution (Part a)

Now we need to calculate the largest possible value of P(X=3)

E(X)=∑xP(X)

1.8=[1×P(X=1)]+[2×P(X=2)]+[3×P(X=3)]

The largest value forP(X=3), the other two probabilitity must containthe smallest value that isP(X=1)=P(X=2)=0

1.8=[1×0]+[2×0]+[3×P(X=3)]

1.8=0+0+3P(X=3)

⇒P(X=3)=1.83

⇒P(X=3)=0.6

03

Final answer (Part a)

The required largest possible value isP(X=3)=0.6.

04

Given information (part b)

Consider 3trials.

X=total number of success

E[X]=1.8

05

Solution (Part b)

Here we need to calculate the lowest possible value of P(X=3)

The probability value always lies between 0and 1. Which shows that the possible smallest values of P(X=3)would be zero(0).

06

Final answer (Part b)

The lowest possible value ofP(X=3)will be0

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