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megankbrown5
30.07.2019 •
Mathematics
Find an equation of the plane that passes through the point and contains the line
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Ответ:
Ответ:
(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) Two diamonds are chosen at random.
The above situation can be represented through Binomial distribution;
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
LET X = Number of diamonds that are industrial dash grade
So, it means X ~![Binom(n=2, p=0.93)](/tpl/images/0572/3870/d6088.png)
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}](/tpl/images/0572/3870/50922.png)
=![1 \times 0.93^{2} \times 1](/tpl/images/0572/3870/c5009.png)
= 0.8649
(b) Six diamonds are chosen at random.
The above situation can be represented through Binomial distribution;
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
LET X = Number of diamonds that are industrial dash grade
So, it means X ~![Binom(n=6, p=0.93)](/tpl/images/0572/3870/2fe65.png)
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}](/tpl/images/0572/3870/98fa2.png)
=![1 \times 0.93^{6} \times 1](/tpl/images/0572/3870/10124.png)
= 0.6469
(c) Here, also 6 diamonds are chosen at random.
The above situation can be represented through Binomial distribution;
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
LET X = Number of diamonds that are of gem dash quality
So, it means X ~![Binom(n=6, p=0.07)](/tpl/images/0572/3870/5e0c7.png)
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}](/tpl/images/0572/3870/c4756.png)
=![1 - [1 \times 1 \times 0.93^{6}]](/tpl/images/0572/3870/8c670.png)
= 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.