<span>When you are sampling from a small finite lot, the hypergeometric distribution applies. The binomial is a poor approximation in this case.
The general equation for the hypergeometric where aCx means the number of combinations of a items selected x at-a-time.:
P(x) =[(aCx)(N-aCn-x)]/NCn
Where
N is the lot size = 20
a is the number of defectives in the lot = 3.
x is the number of defectives in the sample.
n is the sample size = 2.
A. The probability that the first item is defective is
P(x=1) = [(3C1)(17C1)]/(20C2)
= (3)(17)/190 = 0.268
The probability that the second item is defective is
P(x = 1) = [(2C1)(17C1)]/(19C2) = (2)(17)/171 = 0.199.
So the total probability is (0.268)(0.199) = 0.0532
B. The probability that the first item is good is:
P(x = 0) = (3C0)(16C2)]/20C2 = (1)(120)/190 = 0.632
The probability that the second item is defective is
P(x = 0) =[(3C0)(16C2)]/19C2
= (1)(120)/171 = 0.670.
The total probability is 0.632(0.670) = 0.4234</span>
<u>Answer:</u>
The equivalent of -1/3t = 4/3 is t = -4
<u>Solution:</u>
Need to determine such value of t so that when it is multiplied by -1/3, we should get 4/3
That is we need to solve following expression for value of t.
On multiplying both the sides by 3 we get
Now multiplying both the sides by -1 we get
Let’s recheck the value of t
Hence for t = -4, expression -1/3 t is equal to 4/3.
Answer: See explanation
Step-by-step explanation:
Your question isn't complete but let me give different scenarios to help you solve.
Firstly, assuming we want to calculate the amount of items left after the large items are carried alone, this will be:
= 561 - 36
= 525
Secondly, if we want to know the percentage of the large items that will be carried alone. This will be:
= 36/561 × 100.
= 6.42%
Thirdly, if you want to know the percentage of the items that will be left after the large ones have been carried alone. This will be:
= 100% - 6.42%
= 93.58%
