# [SOLVED] Basic probability questions

• Feb 15th 2010, 12:02 AM
downthesun01
[SOLVED] Basic probability questions
I'm studying for an exam and am having a few small problems. I wasn't whether it was best to make a thread for each or post everything in one thread, but in the interest of minimizing clutter I to do the latter.

1.
In a high school graduating class of 100 students, 54 students studied math, 69 studied history, and 35 studied both. If one of these students is selected at random, find the probability that the student took history but not math.

The answers is supposed to be 34/100.
I'm not too sure what I'm looking for. Is this right?

$p(math' \cup history)$

Which would be

$0.46 + 0.69 - p(math' \cap history)=
1.15 - (0.46 * 0.69)=
1.15-0.3174=
0.8326$

Which is obviously wrong.

Looking at the information given. I see that

$p(history) - p(math \cap history) =
0.69 - 0.35 = 0.34$

Which is the right answer but I'm a little confused about it. Looking at it now, it makes sense because you're taking all of the students that took history and removing that section that also took math.

How would one set up the problem using the $\cap$ and $\cup$ symbols? That's the part I'm stuck on.

2.
Interest centers around the life of a particular electrical component. Let A be the event that the component fails a particular test and B be the event that the component displays strain but does not fail. Event A occurs with a probability 0.20 and event B occurs with a probability 0.35. What is the probability that a component works perfectly well (i.e., neither displays strain nor fails the test).

Here's what I did.

$p(A' \cap B') = p(A') * p(B') = 0.8 * 0.65 = 0.52$

Obviously the wrong answer. However, I decided to use De Morgan's law as such

$p(A' \cap B') = p(A \cup B)' = 1 -[0.20 + 0.35] = 1- 0.55 = 0.45$

If $p(A' \cap B') = p(A \cup B)'$ under De Morgan's law, then why did I get two different answers? Where's my mistake?
• Feb 15th 2010, 01:59 AM
mr fantastic
Quote:

Originally Posted by downthesun01
I'm studying for an exam and am having a few small problems. I wasn't whether it was best to make a thread for each or post everything in one thread, but in the interest of minimizing clutter I to do the latter.

1.
In a high school graduating class of 100 students, 54 students studied math, 69 studied history, and 35 studied both. If one of these students is selected at random, find the probability that the student took history but not math.

The answers is supposed to be 34/100.
I'm not too sure what I'm looking for. Is this right?

$p(math' \cup history)$

Which would be

$0.46 + 0.69 - p(math' \cap history)=
1.15 - (0.46 * 0.69)=
1.15-0.3174=
0.8326$

Which is obviously wrong.

Looking at the information given. I see that

$p(history) - p(math \cap history) =
0.69 - 0.35 = 0.34$

Which is the right answer but I'm a little confused about it. Looking at it now, it makes sense because you're taking all of the students that took history and removing that section that also took math.

How would one set up the problem using the $\cap$ and $\cup$ symbols? That's the part I'm stuck on.

2.
Interest centers around the life of a particular electrical component. Let A be the event that the component fails a particular test and B be the event that the component displays strain but does not fail. Event A occurs with a probability 0.20 and event B occurs with a probability 0.35. What is the probability that a component works perfectly well (i.e., neither displays strain nor fails the test).

Here's what I did.

$p(A' \cap B') = p(A') * p(B') = 0.8 * 0.65 = 0.52$

Obviously the wrong answer. However, I decided to use De Morgan's law as such

$p(A' \cap B') = p(A \cup B)' = 1 -[0.20 + 0.35] = 1- 0.55 = 0.45$

If $p(A' \cap B') = p(A \cup B)'$ under De Morgan's law, then why did I get two different answers? Where's my mistake?

For 1, I suggest you draw a Venn diagram.

For 2, it's as simple as 1 - 0.20 - 0.35.
• Feb 15th 2010, 02:03 AM
Hello downthesun01
Quote:

Originally Posted by downthesun01
I'm studying for an exam and am having a few small problems. I wasn't whether it was best to make a thread for each or post everything in one thread, but in the interest of minimizing clutter I to do the latter.

1.
In a high school graduating class of 100 students, 54 students studied math, 69 studied history, and 35 studied both. If one of these students is selected at random, find the probability that the student took history but not math.

The answers is supposed to be 34/100.
I'm not too sure what I'm looking for. Is this right?

$p(math' \cup history)$

Which would be

$0.46 + 0.69 - p(math' \cap history)=
1.15 - (0.46 * 0.69)=
1.15-0.3174=
0.8326$

Which is obviously wrong.

Looking at the information given. I see that

$p(history) - p(math \cap history) =
0.69 - 0.35 = 0.34$

Which is the right answer but I'm a little confused about it. Looking at it now, it makes sense because you're taking all of the students that took history and removing that section that also took math.

How would one set up the problem using the $\cap$ and $\cup$ symbols? That's the part I'm stuck on.

2.
Interest centers around the life of a particular electrical component. Let A be the event that the component fails a particular test and B be the event that the component displays strain but does not fail. Event A occurs with a probability 0.20 and event B occurs with a probability 0.35. What is the probability that a component works perfectly well (i.e., neither displays strain nor fails the test).

Here's what I did.

$p(A' \cap B') = p(A') * p(B') = 0.8 * 0.65 = 0.52$

Obviously the wrong answer. However, I decided to use De Morgan's law as such

$p(A' \cap B') = p(A \cup B)' = 1 -[0.20 + 0.35] = 1- 0.55 = 0.45$

If $p(A' \cap B') = p(A \cup B)'$ under De Morgan's law, then why did I get two different answers? Where's my mistake?

In #1, you're asked for the probability that the student took history but not math; i.e. history and not math. That's $p(\text{history} \cap \text{ math}')$ not $p(\text{history} \cup \text{ math}')$ - that's history or not math.

In #2, you might find it helpful to draw a probability tree representing the possible outcomes from the two events: the component does/does not fail, and the component does/does not show strain.

B is the event that the component shows strain but does not fail. If the probability that it shows strain (whether it fails or not) is $p$, then:
$0.8p = 0.35$
Then you'll see from the tree diagram that the probability that the component does not fail and does not show strain is:
$0.8(1-p) = 0.8 -0.8p = 0.8-0.35=0.45$
The event $A'\cap B'$ is simply not the event you want, because $B'$ is not simply the event that the component doesn't show strain.