Probability question?

2015-09-01 9:01 am
Bag A contains 4 blue balls and 3 red balls. Bag B contains 2 blue balls and 3 red balls. Pierre chooses one ball at random from the 7 balls in bag A and puts it in bag B. He then chooses one ball at random from the 6 balls in bag B. Find the probability he chooses a red ball from bag B.

回答 (3)

2015-09-01 9:10 am
✔ 最佳答案
R: a red ball is drawn
B: a blue ball is drawn

Pr(R from B)
= Pr([R from A] then [R from B]) + Pr([B from A] then [R from B])
= (3/7) × (4/6) + (4/7) × (3/6)
= (12/42) + (12/42)
= 24/42
= 12/21
2015-09-01 9:17 am
There are two scenarios for picking a red ball from bag B.

Red from bag A, then Red from bag B

OR, Blue from bag A, then Red from bag B.

1st case => (3/7) x (4/6) = 12/42

2nd case => (4/7) x (3/6) = 12/42

Hence, 12/42 + 12/42 => 24/42 = 4/7...simplest form

:)>
2015-09-01 9:44 am
3/5 (The Code±1)
1/7 / 3/7 = 1/3 (The Crossover) (3/49 is ultimate probability)
bag stands at 3/5 + 1/3 = 1 4/5

5 balls I think you mean. Maybe you are unbedevilling yourself.

1 4/5
1/5

Then if you 1 5/5 or 1/1 ...

So it's reasonably 50/50 chance or *random* given the technique provided on this occasion. However beginner's luck states that if you want a red ball first time, you'll get it, but after that it will be 50/50 or *random*.

I don't really condone 'probability' because you shouldn't bet away your/our/the money. You need sure things of eternal things. There's possibility & there's probability. The question asked there really is a possibility question. Probable is about having more than 50%. Possible is about having 50%.

Genesis 1:4 states God's luck... because God controls.


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