Q.7.10 - A First Course in Probability

Question

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

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

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

Step-by-Step Solution

Answer

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

Step 1: Given information (Part a)

Consider $3$ trials.

$X=$total number of success

$E\left[X\right]=1.8$

Step 2: Solution (Part a)

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

$E\left(X\right)=\sum xP\left(X\right)$

$1.8=[1\times P(X=1\left)\right]+[2\times P(X=2\left)\right]+[3\times P(X=3\left)\right]$

$\left[\begin{array}{l}\text{ The largest value for }P(X=3)\text{, the other two probabilitity must contain }\\ \text{ the smallest value that is }P(X=1)=P(X=2)=0\end{array}\right]$

$1.8=[1\times 0]+[2\times 0]+[3\times P(X=3\left)\right]$

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

$\Rightarrow P(X=3)=\frac{1.8}{3}$

$\Rightarrow P(X=3)=0.6$

Step 3: Final answer (Part a)

The required largest possible value is $P(X=3)=0.6$.

Step 4: Given information (part b)

Consider $3$ trials.

$X=$ total number of success

$E\left[X\right]=1.8$

Step 5: Solution (Part b)

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

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

Step 6: Final answer (Part b)

The lowest possible value of $P(X=3)$ will be $0$