# Thread: Jar with 5R&2G balls

1. ## Jar with 5R&2G balls

Okay, last one!

You have a jar with 5Red balls & 2Green balls--a ball's chosen randomly. It's returned to the jar along w/6 more of the same color. Then, a ball is drawn again--calculate the probability that the second ball is Red, as a fraction in lowest terms, showing all intermediate steps.

Also, find the prob. that both balls drawn are same color--don't need to simplify this answer.

THANK YOU SO MUCH!!

2. Hello,
hope this is correct..

$\displaystyle Pr(\mbox{Second is red})=P(R|G)+P(R|R)=\frac{2}{7}\cdot \frac{5}{13} + \frac{5}{7}\cdot \frac{11}{13}=\frac{5}{7}$

$\displaystyle Pr(\mbox{Both the same colour})=P(R|R)+P(G|G)=\frac{5}{7}\cdot \frac{11}{13}+\frac{2}{7}\cdot \frac{8}{13}=\frac{71}{91}$

3. Hello, bjanela!

You're right, Robb!

I'll baby-step through it . . .

You have a jar with 5 Red balls and 2 Green balls.
A ball is chosen randomly.
It's returned to the jar along w/6 more of the same color.
Then, a ball is drawn again.

(a) Find the probability that the second ball is Red.
There are two cases to consider:
. . [1] The first ball is Red.
. . [2] The first ball is Green.

Case 1: the first ball is red.
. . $\displaystyle P(\text{1st Red}) \:=\:\frac{5}{7}$

Then the jar has: .11 Red, 2 Green
. . $\displaystyle P(\text{2nd Red}) \:=\:\frac{11}{13}$

Hence: .=$\displaystyle P(\text{1st Red}\:\wedge\:\text{2nd Red}) \:=\:\frac{5}{7}\cdot\frac{11}{13} \:=\:\frac{55}{91}$

Case 2: the first ball is green.
. . $\displaystyle P(\text{1st Green}) \:=\:\frac{2}{7}$

Then the jar has: .5 Red, 8 Green
. . $\displaystyle P(\text{2nd Red}) \:=\:\frac{5}{13}$

Hence: .$\displaystyle P(\text{1st Green}\:\wedge\:\text{2nd Red}) \:=\:\frac{2}{7}\cdot\frac{5}{13} \:=\:\frac{10}{91}$

Therefore: .$\displaystyle P(\text{2nd Red}) \;=\;\frac{55}{91} + \frac{10}{91} \;=\;\frac{65}{91} \;=\;\boxed{\frac{5}{7}}$

(b) Find the probabiity that both balls drawn are same color.

We already know: .$\displaystyle P(\text{1st Red}\:\wedge\:\text{2nd Red}) \;=\;\frac{55}{91}$

We have: .$\displaystyle P(\text{1st Green}) \:=\:\frac{2}{7}$

. . The jar has: .5 Red, 8 Green

. . Then: .$\displaystyle P(\text{2nd Green}) \:=\:\frac{8}{13}$

. . Hence: .$\displaystyle P(\text{1st Green}\:\wedge\:\text{2nd Green}) \:=\:\frac{2}{7}\cdot\frac{8}{13} \:=\:\frac{16}{91}$

Therefore: .$\displaystyle P(\text{same color}) \;=\;\frac{55}{91} + \frac{16}{91} \;=\;\boxed{\frac{71}{91}}$