Answer to Question #130785 in Statistics and Probability for Hon Name

Question

It is known that 90% of a defeated computers can be repaired. A sample of 20 computers is selected randomly. Find the probability that;

1. At least 3 computers can be repaired.

2. 5 or less computers can be repaired.

3. None can be repaired

Based on your output on these how will you advice after a critical analysis from your output

Answer 1

This is a Binomial problem

n=20 and p=0.9

1. $P(\ge3)=1-P(\le2)$

$P(\le2)=P(0)+P(1)+P(2)$

$=\sum_{r=0}^2\binom{20}{r}0.9^r(1-0.9)^{20-r}$

$1E-20+1.8E-18+1.539E-16=1.55712E-16$

Thus, $P(\ge3) = 1-1.55712E-16$ which is approximately equal to one.

2. $P(\le5)=\sum_{r=0}^5\binom{20}{r}0.9^r(1-0.9)^{20-r}$

From part one, $P(\le2)=1.5571E-16$

Thus, $P(\le5)=1.5571E-16+8.3106E-15+3.1788E-139.15496E-12=9.4813E-12$

$P(\le5)=9.4813E-12$ is approximately equal to zero.

3

$P(=0)=\binom{20}{0}(1-0.9)^{20}$

$=1E-20$ from part one above. It is approximately equal to zero.

From the output, it is certain that at least three out of twenty computers can be repaired. From part two and three, it is impossible that only five or less computers out of twenty can be repaired. Therefore there is high change of a computer getting repaired. The advice is, if one has one or more defeated computer(s), he or she should take for repair since there are high chances of getting repaired.