Question

Let A = {1, 2, 3, 4, 5} and B = {2, 4}. What is A ∩ B?

Answer 1

@ganeshie8 @zepdrix

Answer 2

Intersection is asking us, "Hey there are some things in A, which of those are also in B?"
Which elements do they share in common?

Answer 3

2,4?

Answer 4

\[\Large\rm A=\{1,\color{orangered}{2},3,\color{orangered}{4},5\}\]\[\Large\rm B=\{\color{orangered}{2},\color{orangered}{4}\}\]\[\Large\rm A\cap B=\{2,4\}\]Good good good! :)

Answer 5

A Venn diagram is shown below:
What are the elements of (A ∩ B) ' ?

Answer 6

is it 1,7?

Answer 7

So we want the numbers in the region where the circles `intersect`.
Good 1 and 7 are in that intersection!

Answer 8

Ellen plans to have two children but doesn't know if they are boy-boy, girl-girl, or boy-girl. What is the probability that she will have two girls?
0.25
0.5
0.75
1.00

Answer 9

would it be .25 because 1/3 is .333 so you would round .25 to .3?

Answer 10

This is a really weird question.
They should have included girl-boy as one of the options since she can't predict that she'll have a boy first. Kind of sloppy the way they wrote that.

Odds of having a girl for her first birth: \(\Large\rm 1~in~2\)
Odds of having a girl the second time around: \(\Large\rm 1~in~2\)

She wants to calculate the odds of having a girl BOTH times.
So we multiply these values together.

.5 * .5 = .25

You've got the correct answer, but I think the question is a little messed up :p

Answer 11

1 in 2 can be written as a fraction: 1/2 which is equivalent to .5
I hope there wasn't any confusion about that.

Answer 12

no there wasnt :)

Answer 13

Rob determined that some teenagers like to eat hot dogs, some like to eat hamburgers, and others don't like to eat hotdogs or hamburgers. He calculated the probabilities and created the Venn diagram below:
What is the probability that a teenager eats hotdogs, given that he/she eats hamburger?

Answer 14

0.2
0.3
0.33
0.67

Answer 15

Oh boy.. I'm trying to remember probability.
So there's a formula for conditional probability,\[\Large\rm P(A|B)=\frac{P(A\cap B)}{P(B)}\]Does this look familiar?

Answer 16

Hmm that's not leading to the right answer though >.<
Thinking...

Answer 17

no? this is the first time ive ever done this stuff :/

Answer 18

Ok ok here is what I'm thinking.....

Answer 19

Given that we're here