Answer:
(a) 0.8649
(b) 0.6469
(c) 0.353
Step-by-step explanation:
We are given that a diamond can be classified as either gem dash quality or industrial dash grade. Suppose that 93% of diamonds are classified as industrial dash grade.
(a) <u>Two diamonds are chosen at random.</u>
The above situation can be represented through Binomial distribution;
![P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....](https://tex.z-dn.net/?f=P%28X%3Dr%29%20%3D%20%5Cbinom%7Bn%7D%7Br%7Dp%5E%7Br%7D%20%281-p%29%5E%7Bn-r%7D%20%3B%20x%20%3D%200%2C1%2C2%2C3%2C.....)
where, n = number of trials (samples) taken = 2 diamonds
r = number of success = both 2
p = probability of success which in our question is % of diamonds
that are classified as industrial dash grade, i.e; 0.93
<em>LET X = Number of diamonds that are industrial dash grade</em>
So, it means X ~ ![Binom(n=2, p=0.93)](https://tex.z-dn.net/?f=Binom%28n%3D2%2C%20p%3D0.93%29)
Now, Probability that both diamonds are industrial dash grade is given by = P(X = 2)
P(X = 2) = ![\binom{2}{2}\times 0.93^{2} \times (1-0.93)^{2-2}](https://tex.z-dn.net/?f=%5Cbinom%7B2%7D%7B2%7D%5Ctimes%200.93%5E%7B2%7D%20%5Ctimes%20%281-0.93%29%5E%7B2-2%7D)
=
= 0.8649
(b) <u>Six diamonds are chosen at random.</u>
The above situation can be represented through Binomial distribution;
![P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....](https://tex.z-dn.net/?f=P%28X%3Dr%29%20%3D%20%5Cbinom%7Bn%7D%7Br%7Dp%5E%7Br%7D%20%281-p%29%5E%7Bn-r%7D%20%3B%20x%20%3D%200%2C1%2C2%2C3%2C.....)
where, n = number of trials (samples) taken = 6 diamonds
r = number of success = all 6
p = probability of success which in our question is % of diamonds
that are classified as industrial dash grade, i.e; 0.93
<em>LET X = Number of diamonds that are industrial dash grade</em>
So, it means X ~ ![Binom(n=6, p=0.93)](https://tex.z-dn.net/?f=Binom%28n%3D6%2C%20p%3D0.93%29)
Now, Probability that all six diamonds are industrial dash grade is given by = P(X = 6)
P(X = 6) = ![\binom{6}{6}\times 0.93^{6} \times (1-0.93)^{6-6}](https://tex.z-dn.net/?f=%5Cbinom%7B6%7D%7B6%7D%5Ctimes%200.93%5E%7B6%7D%20%5Ctimes%20%281-0.93%29%5E%7B6-6%7D)
=
= 0.6469
(c) <u>Here, also 6 diamonds are chosen at random.</u>
The above situation can be represented through Binomial distribution;
![P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....](https://tex.z-dn.net/?f=P%28X%3Dr%29%20%3D%20%5Cbinom%7Bn%7D%7Br%7Dp%5E%7Br%7D%20%281-p%29%5E%7Bn-r%7D%20%3B%20x%20%3D%200%2C1%2C2%2C3%2C.....)
where, n = number of trials (samples) taken = 6 diamonds
r = number of success = at least one
p = probability of success which is now the % of diamonds
that are classified as gem dash quality, i.e; p = (1 - 0.93) = 0.07
<em>LET X = Number of diamonds that are of gem dash quality</em>
So, it means X ~ ![Binom(n=6, p=0.07)](https://tex.z-dn.net/?f=Binom%28n%3D6%2C%20p%3D0.07%29)
Now, Probability that at least one of six randomly selected diamonds is gem dash quality is given by = P(X
1)
P(X
1) = 1 - P(X = 0)
= ![1 - \binom{6}{0}\times 0.07^{0} \times (1-0.07)^{6-0}](https://tex.z-dn.net/?f=1%20-%20%5Cbinom%7B6%7D%7B0%7D%5Ctimes%200.07%5E%7B0%7D%20%5Ctimes%20%281-0.07%29%5E%7B6-0%7D)
=
= 1 -
= 0.353
Here, the probability that at least one of six randomly selected diamonds is gem dash quality is 0.353 or 35.3%.
For any event to be unusual it's probability is very less such that of less than 5%. Since here the probability is 35.3% which is way higher than 5%.
So, it is not unusual that at least one of six randomly selected diamonds is gem dash quality.