Question

Probability is one of the new chapter for me . and I did't get the concept .

A bag contains 2 red, 3 green and 2 blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue?

A.
10
21
B.
11
21
C.
2
7
D.
5
7

Answer 1

@mayankdevnani @imqwerty

Answer 2

Are the balls replaced after being drawn?

Answer 3

no

Answer 4

from the choices it would seem they are not replaced

Answer 5

Okay, so let's start with the first ball. The bag currently has 7 balls in it, of which 2 are blue, so the probability of drawing a blue ball is \(\frac{2}{7}\). Do you follow so far?

Answer 6

yeah! got it

Answer 7

Okay, now can you tell me the probability of the second ball taken being blue? Remember that the bag is now missing one blue ball.

Answer 8

1/6 ?

Answer 9

Exactly, good work. Now we have the probability of two events happening, do you know how to work out the probability of both happening?

Answer 10

no idea :(

Answer 11

Oh crap, completely misread your question. Ignore that. Let's wind back a bit here.

Answer 12

:O ok

Answer 13

I read it as both blue when it's neither blue. Okay, well now you have a little understanding about probability, can you tell me the probability of the first ball NOT being blue?

Answer 14

wild guess !!
5/7 ?

Answer 15

Exactly, well done. There are 5 balls in the bag that aren't blue, out of the 7. What about the second ball?

Answer 16

4/6 ?

Answer 17

Yep, after removing one of the non-blue balls we're left with 6 balls in total, 4 of which aren't blue. Can you read this and see if you can tell me the probability of both happening:

http://www.bbc.co.uk/schools/gcsebitesize/maths/statistics/probabilityhirev2.shtml

Answer 18

i think it should be ,
1/7 * 1/6
= 1/42

Answer 19

You're correct about the multiplication, P(A and B)=P(A)P(B) but your answer is wrong -this is partly my fault for me confusing you. You've used the wrong values, it should be (5/7)*(4/6).

Answer 20

:( did't get this

Answer 21

1/7 and 1/6 were the probabilities we worked out before for drawing a blue ball then a blue ball - this wasn't the right thing to work out and is my fault for misreading the question. We then worked out the probability of the first ball being NOT blue as 5/7, and then we worked out the probability of the second ball being NOT blue as 4/6. So the probability of these two events both happening is (5/7)*(4/6). Do you understand now?

Answer 22

oh ! yeah got it .

Answer 23

thanks for being patient and teach me @tom982 :)

Answer 24

another way of doing this wouild be to draw a tree diagran. Have you faniliar with them?

Answer 25

nope

Answer 26

you probably be introduced to them sometime

Answer 27

:)

Answer 28

No problem. I was tempted to do the tree diagram @welshfella but wasn't sure if it had been covered yet.