Spaceman Spiff’s spacecraft has a warning light that is supposed to switch on when the freem

blasters are overheated. Let W be the event “the warning light is switched on” and F “the freem
blasters are overheated”. Suppose the probability of freem blaster overheating P(F) = 0.1 and
that the light is switched on when they actually are overheated is 0.99, and that there is a 2%
chance that it comes on when nothing is wrong: P(W|F
c
) = 0.02. Use a probability tree to
support your answer. Determine the probability that the warning light is switched on. Determine
the conditional probability that the freem blasters are overheated, given that the warning light is on

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To determine the probability that the warning light is switched on, we can use the law of total probability and the information given:

Let's denote:
- F as the event "the freem blasters are overheated"
- W as the event "the warning light is switched on"
- P(Fc) as the probability of F not occurring (complement of F), which is equal to 1 - P(F)

We are given:
- P(F) = 0.1 (probability of freem blasters overheating)
- P(W|F) = 0.99 (probability of the warning light being switched on, given that freem blasters are overheated)
- P(W|Fc) = 0.02 (probability of the warning light being switched on, given that freem blasters are not overheated)

Using a probability tree, we can visualize the information:

|--------- F ---------| |-------- Fc --------|
| | | |
| | | |
W |---| 0.99 | |---| 0.02 |
| | | |
|--------- W ---------| |-------- W --------|
| | | |
| | | |
F |---| 0.1 | |---| 0.9 |

To find the probability that the warning light is switched on (P(W)), we can apply the law of total probability:
P(W) = P(W|F) * P(F) + P(W|Fc) * P(Fc)
= 0.99 * 0.1 + 0.02 * 0.9
= 0.099 + 0.018
= 0.117

Therefore, the probability that the warning light is switched on is 0.117.

To find the conditional probability that the freem blasters are overheated, given that the warning light is on (P(F|W)), we can use Bayes' theorem:
P(F|W) = P(W|F) * P(F) / P(W)
= 0.99 * 0.1 / 0.117
= 0.099 / 0.117
≈ 0.8462

Therefore, the conditional probability that the freem blasters are overheated, given that the warning light is on, is approximately 0.8462.

To determine the probability that the warning light is switched on, we can use the law of total probability. We will consider two cases: when the freem blasters are actually overheated (F) and when they are not overheated (~F).

Case 1: Freem blasters are overheated (F)
Given:
P(F) = 0.1 (probability of freem blasters overheating)
P(W|F) = 0.99 (probability of the warning light being switched on when freem blasters are overheated)

So, the probability of the warning light being switched on when the freem blasters are overheated is P(W and F) = P(W|F) * P(F) = 0.99 * 0.1 = 0.099.

Case 2: Freem blasters are not overheated (~F)
Given:
P(~F) = 1 - P(F) = 1 - 0.1 = 0.9 (probability of freem blasters not overheating)
P(W|~F) = 0.02 (probability of the warning light being switched on when freem blasters are not overheated)

So, the probability of the warning light being switched on when the freem blasters are not overheated is P(W and ~F) = P(W|~F) * P(~F) = 0.02 * 0.9 = 0.018.

Now, we can use the law of total probability to find the probability of the warning light being switched on:
P(W) = P(W and F) + P(W and ~F) = 0.099 + 0.018 = 0.117.

Therefore, the probability that the warning light is switched on is 0.117.

To determine the conditional probability that the freem blasters are overheated, given that the warning light is on, we can use Bayes' theorem:

P(F|W) = (P(W|F) * P(F)) / P(W) = (0.99 * 0.1) / 0.117 ≈ 0.846.

Therefore, the conditional probability that the freem blasters are overheated, given that the warning light is on, is approximately 0.846.