Question

Which equation implies that A and B are independent events?

A. P(A U B) = P(A)/P(B)
B. P(A) = P(B)
C. P(B|A) = P(A)
D. P(A U B) = P(A) x P (B)
E. P(A|B) = P(B|A)

Answer 1

if A and B are independent events then
P(B|A) = P(A)

Answer 2

@sidsiddhartha that can't be right, since
\[P(B|A)=\frac{P(B\cap A)}{P(A)}=\frac{P(B)P(A)}{P(A)}=P(B)\]
where \(P(B\cap A)=P(B)P(A)\) because they're independent events.

Answer 3

So that leaves me with E?

Answer 4

No, I actually don't think any of these choices are right.

From E, you have
\[\begin{align*}P(A|B)&=P(B|A)\\\\
\frac{P(A\cap B)}{P(B)}&=\frac{P(B\cap A)}{P(A)}\\\\
P(A)&=P(B)
\end{align*}\]
which doesn't tell you anything about independence. This is the same as option B, so we can eliminate that one too.

The closest one is D, but it says \(P(A\cup B)\), which is not the same as \(P(A\cap B)\). Are you sure that's supposed to be a union \(\cup\) and not an intersection \(\cap\) ?

Answer 5

Well there you go, it's D. For independent events \(A\) and \(B\), you have
\[P(A\cap B)=P(A)\times P(B)\]

Answer 6

truee thanks :) can you help me with something else..

Answer 7

@SithsAndGiggles

Answer 8

sure

Answer 9

Which of the following demonstrates a fair choice of 1 from among 12 candidates?

Answer 10

You want the one that has 12 equally likely outcomes. The only one that does that seems to be the first option. A die has 6 sides and a coin has 2, so you have 12 total outcomes (1-h, 2-h, 3-h, and so on).

Answer 11

The second and third have 16 outcomes and the last two have 36 outcomes, so it's definitely the first one.

Answer 12

sorry i made a mistake there
it should be P(A/B) = P(AB)/P(B)=P(A)

Answer 13

how about this other one. sorry but my brain is not working well today

Answer 14

You toss 5 coins. What is the probability that all 5 show heads?

1/8
1/32
1/16
1/12
1/4