Answer: 0.5898
Step-by-step explanation:
Given : J.J. Redick of the Los Angeles Clippers had a free throw shooting percentage of 0.901 .
We assume that,
The probability that .J. Redick makes any given free throw =0.901 (1)
Free throws are independent.
So it is a binomial distribution .
Using binomial probability formula, the probability of getting success in x trials :
, where n= total trials
p= probability of getting in each trial.
Let x be binomial variable that represents the number of a=makes.
n= 14
p= 0.901 (from (1))
The probability that he makes at least 13 of them will be :-
![=^{14}C_{13}(0.901)^{13}(1-0.901)^1+^{14}C_{14}(0.901)^{14}(1-0.901)^0\\\\=(14)(0.901)^{13}(0.099)+(1)(0.901)^{14}\ \ [\because\ ^nC_n=1\ \&\ ^nC_{n-1}=n ]\\\\\approx0.3574+0.2324=0.5898](https://tex.z-dn.net/?f=%3D%5E%7B14%7DC_%7B13%7D%280.901%29%5E%7B13%7D%281-0.901%29%5E1%2B%5E%7B14%7DC_%7B14%7D%280.901%29%5E%7B14%7D%281-0.901%29%5E0%5C%5C%5C%5C%3D%2814%29%280.901%29%5E%7B13%7D%280.099%29%2B%281%29%280.901%29%5E%7B14%7D%5C%20%5C%20%5B%5Cbecause%5C%20%5EnC_n%3D1%5C%20%5C%26%5C%20%5EnC_%7Bn-1%7D%3Dn%20%5D%5C%5C%5C%5C%5Capprox0.3574%2B0.2324%3D0.5898)
∴ The required probability = 0.5898
The lines y=x+2 and y=x are parallel, because they have the same slope.
(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.
We are given that each share will receive a dividend equal
to 56.25. In this problem, we should have been given the total number of shares
that PRH has so that we can know the dividend. Anyway, the formula to calculate
the dividend is:
<span>Dividend = 56.25 * (Number of Shares)</span>
He soled 21 and 1/2 of almonds at the fair because 25 is = to 24 and 2/2 so that - 3 and 1/2 is = to 21 and 1/2
