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Exercise 9 Page 719
a Are the events independent? What can you say about the probability of three independent events occurring at the same time?
b What event is the complement of at least one of the mowers being unusable? Can you find its probability?
c The least-reliable mower is the one with the greatest chance of being unusable.
a 0.1 %
b 32.1 %
c The chances of the business being productive get smaller.
Practice makes perfect
a The owner of a lawn-mowing business has three mowers. Only if at least one of the mowers is working, the owner can stay productive. Here is what we know about the mowers.
| Description | |
|---|---|
| Mower I | unusable 10 % of the time |
| Mower II | unusable 8 % of the time |
| Mower III | unusable 18 % of the time |
P( A and B and C ) = P(A) * P(B) * P(C) Since the mowers work independently, the events of any of them breaking down are also independent. Now, if a mower is unusable 10 % of the time, the probability of it being unusable equals 0.1. We can create a table showing the probabilities for all three mowers.
| Probability of Being Unusable | |
|---|---|
| Mower I | 0.1 |
| Mower II | 0.08 |
| Mower III | 0.18 |
Since these events are independent, the probability of all the mowers being unusable at once equals the product of the above probabilities. ( 0.1)( 0.08)( 0.18) Let's evaluate the product. ( 0.1)( 0.08)( 0.18) &= 0.00144 & ≈ 0.001 The probability that all three mowers fail at the same time equals approximately 0.001, or 0.1 %.
b We want to find the probability that at least one of the mowers is unusable on a given day. For that, notice that the following two events are complementary.
| Event A | At least one of the mowers is unusable |
|---|---|
| Event A | All of the mowers are usable |
Before we do this, though, we need to find the probability of each of the mowers working.
| Probability of Being Unusable | Probability of Being Usable | |
|---|---|---|
| Mower I | 0.1 | 1 - 0.1 = 0.9 |
| Mower II | 0.08 | 1- 0.08=0.92 |
| Mower III | 0.18 | 1- 0.18=0.82 |
Now, since the events are independent, we can multiply the probabilities. This will give us the probability P(A) of all the mowers working at once. P(A) &= (0.9)(0.92)(0.82) & ≈ 0.679 Since we know that P(A) and P(A) add up to 1, after substituting P(A) we get the following. P(A) + 0.679 = 1 We can now solve for P(A). P(A) &= 1- 0.679 &= 0.321 The probability of at least one mower being unusable on a given day is approximately 0.321=32.1 %.
c The least-reliable mower stops working completely. The least-reliable mower is the one with the greatest chance of being unusable, which in this case is the third one.
| Probability of Being Unusable | |
|---|---|
| Mower I | 0.1 |
| Mower II | 0.08 |
| Mower III | 0.18 |
We want to find out how this affects the productivity of the business. First, notice that the owner cannot work only if the two remaining mowers are both unusable at the same time. Since the events are independent, the chances of that happening can be calculated as follows. (0.1)(0.08)=0.008 If the least-reliable mower breaks down permanently, the probability that the remaining two mowers are unavailable at the same time on a given day is about 0.008, or 0.8 %. In Part A we found that the probability of all three machines not working at once equals approximately 0.001, which is less than 0.008. 0.008 > 0.001
We see that when the least reliable mower stops working, the probability that the owner's day is unproductive rises.Loading content