Original equation is 
So, 
and
If we compare this equation with the given options, we can easily find that this matches with the last one 
with P = p/2.
Hence, correct option is .
Answer: 
Step-by-step explanation:
Given the following expression:
You need to substitute the given values of "a" and "b" into the expression. Notice that these values are:
Then;
Now you must solve the multiplications:
The final step is to add the numbers. Therefore, you get the following answer:
(a) 0.059582148 probability of exactly 3 defective out of 20
(b) 0.98598125 probability that at least 5 need to be tested to find 2 defective.
(a) For exactly 3 defective computers, we need to find the calculate the probability of 3 defective computers with 17 good computers, and then multiply by the number of ways we could arrange those computers. So
0.05^3 * (1 - 0.05)^(20-3) * 20! / (3!(20-3)!)
= 0.05^3 * 0.95^17 * 20! / (3!17!)
= 0.05^3 * 0.95^17 * 20*19*18*17! / (3!17!)
= 0.05^3 * 0.95^17 * 20*19*18 / (1*2*3)
= 0.05^3 * 0.95^17 * 20*19*(2*3*3) / (2*3)
= 0.05^3 * 0.95^17 * 20*19*3
= 0.000125* 0.418120335 * 1140
= 0.059582148
(b) For this problem, let's recast the problem into "What's the probability of having only 0 or 1 defective computers out of 4?" After all, if at most 1 defective computers have been found, then a fifth computer would need to be tested in order to attempt to find another defective computer. So the probability of getting 0 defective computers out of 4 is (1-0.05)^4 = 0.95^4 = 0.81450625.
The probability of getting exactly 1 defective computer out of 4 is 0.05*(1-0.05)^3*4!/(1!(4-1)!)
= 0.05*0.95^3*24/(1!3!)
= 0.05*0.857375*24/6
= 0.171475
So the probability of getting only 0 or 1 defective computers out of the 1st 4 is 0.81450625 + 0.171475 = 0.98598125 which is also the probability that at least 5 computers need to be tested.
Answer:
a) The expected number of questions that are answered correctly by both A and B = 11 (7 + 4).
b) The Variance of the number of questions that are answered correctly by either A or B = 2.25.
Step-by-step explanation:
Number of questions in the examination = 10
Probability of A's answer being correct = 0.7
Probability of B's answer being correct = 0.4
The expected number of questions that are answered correctly by both A and B:
Probability of Expected
Correct Answer Value Variance
A 0.7 7 (0.7 * 10) 2.25
B 0.4 4 (0.4 * 10) 2.25
Total expected value = 11
Mean = 5.5 2.25
Answer:1. C 3. 52
Step-by-step explanation:
