(a+b)^7= a^7+ 7a^7b+ 21 a^6b²+ 35a^5b³+ 35 a⁴b⁴+ 21 a³b^5 + 7a²b^6 + b^7
What is the probability of picking a number less than 15 for my jar with papers labeled from 1 to 12
1 answer:
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Answer:
(a) Probability that a randomly chosen item is defective and cannot be repaired is 8%.
(b) Probability that exactly 2 of 20 randomly chosen items are defective and cannot be repaired is 0.2711.
Step-by-step explanation:
We are given that of the items manufactured by a certain process, 20% are defective. Of the defective items, 60% can be repaired.
Let Probability that item are defective = P(D) = 0.20
Also, R = event of item being repaired
Probability of items being repaired from the given defective items = P(R/D) = 0.60
<em>So, Probability of items not being repaired from the given defective items = P(R'/D) = 1 - P(R/D) = 1 - 0.60 = 0.40 </em>
(a) Probability that a randomly chosen item is defective and cannot be repaired = Probability of items being defective Probability of items not being repaired from the given defective items
= 0.20 0.40 = 0.08 or 8%
So, probability that a randomly chosen item is defective and cannot be repaired is 8%.
(b) Now we have to find the probability that exactly 2 of 20 randomly chosen items are defective and cannot be repaired.
The above situation can be represented through Binomial distribution;
where, n = number of trials (samples) taken = 20 items
r = number of success = exactly 2
p = probability of success which in our question is % of randomly
chosen item to be defective and cannot be repaired, i.e; 8%
<em>LET X = Number of items that are defective and cannot be repaired</em>
So, it means X ~
Now, Probability that exactly 2 of 20 randomly chosen items are defective and cannot be repaired is given by = P(X = 2)
P(X = 2) =
=
= 0.2711
<em>Therefore, probability that exactly 2 of 20 randomly chosen items are defective and cannot be repaired is </em><em>0.2711.</em>