Jen figures she has a 70% chance of passing math and an 80% chance of passing history

a) What is the probability that she passes math and history? What assumption did you have to make to answer this question?

b) What is the probability that she will pass one but not both? Note: There are two different ways that this can happen. Find the probability of both.

c) What is the possibility that she will pass neither?

d) Is there any other possible outcome for Jenn, other than those listed in a, b and c? What should the three probabilities in a, b and c add up to? Do your probabilities add up correctly?

If the events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events.

Either-or probabilities are found by adding the individual probabilities.

I will do one problem for you, so you can see the process.

b) .7 * (1-.8) = .7 * .2 = .14

.8 * (1-.7) = .8 * .3 = .24

.14 + .24 = .38

a) To find the probability that Jen passes both math and history, we need to assume that the events are independent. This means that the outcome of one event does not affect the outcome of the other event.

The probability of passing math is 70% (0.70), and the probability of passing history is 80% (0.80). To calculate the probability of both events occurring, we multiply the probabilities:

Probability of passing both = 0.70 * 0.80 = 0.56 or 56%

b) There are two ways for Jen to pass one subject but not the other:
1) Pass math and fail history
2) Fail math and pass history

To find the probability of passing one but not both, we add these two probabilities together:

Probability of passing one but not both = Probability of passing math and failing history + Probability of failing math and passing history

Probability of passing one but not both = (0.70 * 0.20) + (0.30 * 0.80) = 0.14 + 0.24 = 0.38 or 38%

c) The probability that Jen will pass neither math nor history is the complement of passing both:

Probability of passing neither = 1 - Probability of passing both = 1 - 0.56 = 0.44 or 44%

d) The other possible outcome for Jen is passing both math and history plus passing one but not both. So the three probabilities in a, b, and c should add up to 1:

0.56 + 0.38 + 0.44 = 1

Yes, the probabilities add up correctly to 1.

a) To find the probability that Jen passes both math and history, we need to multiply the individual probabilities. Thus, the probability that she passes math and history is 70% * 80% = 56%.

To answer this question, we assume that the probability of passing math and history are independent events. In other words, the outcome of one does not influence the outcome of the other. Additionally, we assume that Jen's chances of passing each subject remain constant and do not change over time.

b) There are two ways for Jen to pass one but not both subjects:
i) Pass math but fail history: Probability of passing math * Probability of failing history
= 70% * (1 - 80%) = 70% * 20% = 14%.
ii) Fail math but pass history: Probability of failing math * Probability of passing history
= (1 - 70%) * 80% = 30% * 80% = 24%.

Therefore, the probability that she will pass one but not both subjects is 14% + 24% = 38%.

c) The probability that she will pass neither subject can be calculated by subtracting the sum of the probabilities of all possible outcomes (passing math, passing history, passing both) from 100%.
Probability of not passing either subject = 100% - (70% + 80% - 56%) = 100% - 94% = 6%.

d) No, there is no other possible outcome for Jen as the question only asks about passing or failing math and history. The three probabilities in parts a), b), and c) should add up to 100% since they encompass all the possibilities. In this case, 56% + 38% + 6% = 100%, so the probabilities do add up correctly.

a) To find the probability that Jen passes both math and history, we need to multiply the probabilities of each event happening. Since the events are independent (the outcome of one does not affect the outcome of the other), we can multiply the probabilities:

P(passing math and history) = P(passing math) × P(passing history) = 0.7 × 0.8 = 0.56

The assumption made here is that Jen's chances of passing math and history are independent of each other.

b) To find the probability that she will pass one but not both, there are two different ways this can happen: either she passes math but fails history, or she passes history but fails math.

P(passing math and failing history) = P(passing math) × P(failing history) = 0.7 × (1 - 0.8) = 0.7 × 0.2 = 0.14

P(failing math and passing history) = P(failing math) × P(passing history) = (1 - 0.7) × 0.8 = 0.3 × 0.8 = 0.24

To get the probability of either of these events happening (passing one but not both), we add these probabilities together:

P(passing one but not both) = P(passing math and failing history) + P(failing math and passing history) = 0.14 + 0.24 = 0.38

c) The possibility that she will pass neither math nor history (failing both) can be found by multiplying the probabilities of failing math and failing history:

P(failing both) = P(failing math) × P(failing history) = (1 - 0.7) × (1 - 0.8) = 0.3 × 0.2 = 0.06

d) There is another possible outcome for Jen, which is passing both math and history. So the four possible outcomes for Jen are:
1. Passing math and history
2. Passing math but failing history
3. Failing math but passing history
4. Failing both math and history

The sum of the probabilities for all these outcomes should equal 1, as one of these outcomes is guaranteed to happen. Let's check if the probabilities add up correctly:

0.56 + 0.14 + 0.24 + 0.06 = 1

Yes, the probabilities add up correctly.