From a shipment of 65 transistors, 4 of which are defective, a sample of 6 transistors is selected at random.
(a) In how many different ways can the sample be selected?
ways
(b) How many samples contain exactly 3 defective transistors?
166320 samples
(c) How many samples do not contain any defective transistors?
samples
Hypergeometric distribution applies to a limited population from which a sample is drawn without replacement.
Here we assume all defectives are identical, so are undefectives.
D=4 (defective)
U=65-4=61 (undefective) =>
S=U+D=65=size of population=65
d=defectives selected
u=undefectives selected =>
s=u+d=size of sample=6
C(n,r)=n!/(r!(n-r)!)=number of possible combinations selecting r from n objects.
(a)
Number of ways
=C(S,s)
=C(65,6)
=82598880
(b)
u=3, d=3
Number of samples with exactly 3 defectives
=C(D,d)*C(U,u)
=C(4,3)*C(61,3)
=143960
(c)
u=6
d=0
Number of samples without defectives
=C(D,d)*C(U,u)
=C(4,0)*C(61,6)
=55525372
From a shipment of 70 transistors, 4 of which are defective, a sample of 5 transistors is selected at random.
(a) In how many different ways can the sample be selected?
There are 65 transistors and 6 need to be selected without replacement. This can be calculated using the formula for combinations, which is nCr = n! / (r!(n-r)!), where n is the total number of items and r is the number of items to be selected.
So, in this case, it would be 65C6 = 65! / (6!(65-6)!) = 65! / (6!59!) = 65 * 64 * 63 * 62 * 61 * 60 / (6*5*4*3*2*1) = 814,506,000 different ways.
(b) How many samples contain exactly 3 defective transistors?
Since there are 4 defective transistors, the number of ways to choose 3 of them is 4C3 = 4! / (3!(4-3)!) = 4.
Once we have chosen the 3 defective transistors, we need to choose the remaining 3 non-defective ones from the remaining 61 transistors. So, it would be 61C3 = 61! / (3!(61-3)!) = 61! / (3!58!) = 21,700.
Therefore, the number of samples with exactly 3 defective transistors is 4 * 21,700 = 86,800 samples.
(c) How many samples do not contain any defective transistors?
Since there are 4 defective transistors, we need to choose all 6 non-defective ones from the remaining 61 transistors. So, it would be 61C6 = 61! / (6!(61-6)!) = 61! / (6!55!) = 355,775.
Therefore, the number of samples without any defective transistors is 355,775 samples.
(a) To calculate the number of ways the sample can be selected, we can use the combination formula. The combination formula is given by nCr = n! / (r! * (n-r)!), where n is the total number of items and r is the number of items in each sample.
In this case, we have 65 transistors and we are selecting a sample of 6 transistors. Therefore, the number of ways to select the sample can be calculated as 65C6 = 65! / (6! * (65-6)!).
Calculating this value gives us 65! / (6! * 59!) = (65 * 64 * 63 * 62 * 61 * 60) / (6 * 5 * 4 * 3 * 2 * 1) = 65,780 different ways.
Therefore, there are 65,780 different ways the sample can be selected.
(b) To calculate the number of samples containing exactly 3 defective transistors, we need to consider the number of ways to choose 3 defective transistors from the 4 defective ones, and also the number of ways to choose the remaining 3 transistors from the remaining 61 non-defective ones.
The number of ways to choose 3 defective transistors from 4 can be calculated using the combination formula as 4C3 = 4! / (3! * (4-3)!) = 4.
The number of ways to choose the remaining 3 non-defective transistors from the 61 non-defective ones can be calculated as 61C3 = 61! / (3! * (61-3)!) = (61 * 60 * 59) / (3 * 2 * 1) = 21,220.
To calculate the number of samples containing exactly 3 defective transistors, we need to multiply the number of ways to choose 3 defective transistors with the number of ways to choose 3 non-defective transistors.
Therefore, the number of samples containing exactly 3 defective transistors is 4 * 21,220 = 84,880.
(c) To calculate the number of samples that do not contain any defective transistors, we need to choose all 6 transistors from the 61 non-defective ones.
The number of ways to choose all 6 transistors from the 61 non-defective ones can be calculated as 61C6 = 61! / (6! * (61-6)!) = (61 * 60 * 59 * 58 * 57 * 56) / (6 * 5 * 4 * 3 * 2 * 1) = 5,782,368.
Therefore, the number of samples that do not contain any defective transistors is 5,782,368.