# Really hard probabilty question. HELP!!!!

There are some red counters and some white counters in a bag. At the start, 7 of the counters are white.
Alfie takes at random a counter from the bag.
He does not put the counter back in the bag.

The probability that the first counter Alfie takes is white AND the second counter Alfie takes is 21/80
Work out the number of white counters in the bag at the start.
************* btw the answer is 9
And it involves some algebra
PLEEEAAAAASSSEEEEE HELP
Is this a troll? you just said there were 7 white counters.
Original post by jbman690
Is this a troll? you just said there were 7 white counters.

Lol, I don't think so. What the OP hasn't told you is that two of the "red" counters are actually white. They have just been painted red on. Good question, really gets the mind working.

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Original post by TrashyJac
There are some red counters and some white counters in a bag. At the start, 7 of the counters are white.
Alfie takes at random a counter from the bag.
He does not put the counter back in the bag.

The probability that the first counter Alfie takes is white AND the second counter Alfie takes is 21/80
Work out the number of white counters in the bag at the start.
************* btw the answer is 9
And it involves some algebra
PLEEEAAAAASSSEEEEE HELP

i presume you mean - how many red counters are there? Also what colour is the second counter - red I think, given the answer.

if there were n red counters what is the probability (in terms of n) of taking out white then red? use your answer to set up an equation in n.
If in doubt, draw a tree. The four branches of the tree are WW, WR, RW and RR, but we're told that the first counter chosen is white so we only need to bother with the WW and WR branches.

At the start, there are 7 white counters and n red counters.

For the first choice, P(W) = 7/(n + 7). Can you continue from there to find P(WW) and P(WR)?

After that, you need to look further at P(WW) and P(WR) to see which of them yields an integer solution for n. (You will be solving quadratics in n).
Original post by old_engineer
If in doubt, draw a tree. The four branches of the tree are WW, WR, RW and RR, but we're told that the first counter chosen is white so we only need to bother with the WW and WR branches.

At the start, there are 7 white counters and n red counters.

For the first choice, P(W) = 7/(n + 7). Can you continue from there to find P(WW) and P(WR)?

After that, you need to look further at P(WW) and P(WR) to see which of them yields an integer solution for n. (You will be solving quadratics in n).

Thank you sooooo much I got the answer now
P(first counter is white)= w/(7+w)
P(second counter is red)=7/(7+w-1)

w stands for white

we know w/(7+w) x 7/(7+w-1) is 21/80

through trial and error it is easy to work out that when w = 9 you get 63/240 which you can simplify down to 21/80 by dividing the numerator and demonimator by 3.
what did you get for the answer?
Original post by silly_goose_666
what did you get for the answer?

And you can see that the answer is actually given in the post before yours
Original post by davros