Question

There are 4 blue disks and 3 yellow disks in a bag. Jane and Mary play a game where they each have a turn at drawing a disc with replacement. Jane starts.

Jane wins when she draws a blue disc and Mary wins when she draws a yellow disc.

(a) What's the probability that Jane wins on the first draw?

(b) What is the probability that Jane wins in the first 3 draws?

(c) What is the probability that Jane wins the game?

Answer 1

ready?

Answer 2

yes, how do i start?

Answer 3

first question is easy enough
jane starts
there are 4 blue disks and 7 total, so the probability she wins on the first draw is \(\frac{4}{7}\)
is that part ok?

Answer 4

yes, what about the second part?

Answer 5

is the next part \[\frac{4}{7} *\frac{3}{6} *\frac{2}{5} \]?

Answer 6

ok now we want the probability jane wins in the first 3 draws, which i take to mean either her first draw, her second draw, or her third draw
we have already computed the probability she wins on her first
now lets compute the probability she wins on her second
it is going to take more work
don't forget this is with replacement

Answer 7

how could she win on her second draw?
jane draws yellow, \(p=\frac{3}{7}\) then
mary draws blue \(p=\frac{4}{7}\) then
jane draws blue \(p=\frac{4}{7}\)

Answer 8

hope it is clear what the logic is and where i got the numbers from
in other words it goes
jane doesn't win
mary doesn't win
jane does win
that probability is the product of the above numbers, since they all have to occur, and that product is \((\frac{4}{7})^2\times \frac{3}{7}\)

Answer 9

now you have to do it one more time
jane doesn't win
mary doesn't win
jane doesn't win
mary does't win
jane DOES win

compute each of these probabilities (they are not hard, remember it is with replacement) and then multiply them together
use exponential notation and do not actually compute
i can write it out for you if you like

don't despair, the last question is the easiest

Answer 10

did you compute the last one? i think that problem is there to get you to see the pattern and have you sum a geometric series for the last problem
we can do it a snap way if you like

Answer 11

is the second one\[\frac{4}{7} \left(\frac{3}{4}*\frac{4}{7}*\frac{4}{7}\right) +\frac{3}{7} *\frac{4}{7} *\frac{3}{7} *\frac{4}{7} *\frac{4}{7} \]?
sorry im horrible with probability

Answer 12

looks good but lets write in exponential notation

Answer 13

\[\frac{4}{7} + \left(\frac{3}{4}*\frac{4}{7}*\frac{4}{7}\right) +\frac{3}{7} *\frac{4}{7} *\frac{3}{7} *\frac{4}{7} *\frac{4}{7}\] is more like it

Answer 14

the first \(\frac{4}{7}\) for win on the first try

Answer 15

clear so far?

Answer 16

oh yeah woops

Answer 17

yes, how do i do the next one?

Answer 18

now lets write it so we see the pattern

Answer 19

looks like \[\frac{4}{7}+\frac{4}{7}\times \left(\frac{4}{7}\times \frac{3}{7}\right)+\frac{4}{7}\times \left(\frac{4}{7}\times \frac{3}{7}\right)^2\]

Answer 20

pattern is clear yes? and it will continue
you can add this up if you know how to add a geometric series
but you can solve the problem a different way as well

Answer 21

yeah, its a GP. what is the different way that i can solve it?

Answer 22

we think
actually this is rather interesting because it is the probabilistic proof of how to sum a geometric series
we are only interested in the probability than jane wins
so lets put that as \(J\)

Answer 23

what could happen?
she could win on the first try, and that happens with probability \(\frac{4}{7}\)

Answer 24

OR she could not win, mary not win. this happens with probability \(\frac{3}{7}\times \frac{4}{7}\) and then it is jane's turn again
now her probability of not winning is identical to what it was on her first try

Answer 25

so we get the equation for \(J\) as
\[J=\frac{4}{7}+\frac{4}{7}\times \frac{3}{7}J\]

Answer 26

i hope the reasoning is clear, why the addition and why the multiplication now if you do the algebra you will see solving for J is identical to summing the geometric series written above

Answer 27

i got it thanks!

Answer 28

for the first one you are going to write \[\frac{\frac{4}{7}}{1-\frac{3}{7}\times \frac{4}{7}}\] and you are going to solve the linear equation in exactly the same fashion

Answer 29

yw