Question

Consider 4 trials, each having the same probability of success. Let X denote the total number of successes in these trials.

If E[X]=0.8, find each of the following.

(a) The largest possible value of P{X=4}

(b) The smallest possible value of P{X=4}.

In this case, give possible values for the remaining probabilities:

P{X=0};
P{X=1};
P{X=2};
P{X=3};

Answer 1

Answer:

a) 0.2

b) 0

Step-by-step explanation:

E[X] = 0*P(X=0)+ 1*P(X=1)+2*P(X=2)+3*P(X=3)+4*P(X=4) = 0.8

(a) The largest possible value of P(X=4) occurs when

P(X=1) = P(X=2) = P(X=3) = 0.

In that case,  

4*P(X=4) = 0.8 ===> P(X=4) = 0.8/4 = 0.2

(b) The smallest possible value of P(X=4).

Obviously, the smallest possible value is P(X=4) = 0

In that case, possible values for the remaining probabilities

could be, for example

P(X=0) = P(X=2) = P(X=3) = 0

 

P(X=1) = 0.8

Answer 2

Probabilities are used to determine the chances of events

1. The largest value of P(X = 4) is 0.2
2. The smallest vale of P(x = 4) is 0.

The given parameters are:

$n= 4$

$E(x) = 0.8$

The expected value of a discrete probability distribution is calculated as:

$E(x) = \sum x * P(x)$

So, we have:

$E(X) = 0*P(X=0)+ 1*P(X=1)+2*P(X=2)+3*P(X=3)+4*P(X=4)$

The equation becomes

$0*P(X=0)+ 1*P(X=1)+2*P(X=2)+3*P(X=3)+4*P(X=4) = 0.8$

This gives

$P(X=1)+2P(X=2)+3P(X=3)+4P(X=4) = 0.8$

The above shows that we can make the following assumption

$P(x) = 0$

The largest value of P(X = 4) is at:

$P(X=1) = P(X=2) = P(X=3) = 0.$

So, we have:

$4P(X=4) = 0.8$

Divide both sides by 4

$P(X=4) = 0.2$

The smallest vale of P(x = 4) is 0.

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