Dependable events probabilities

[Zelda22]

Box 1 (RRGG) Box 2 (RRRGG)
A ball is drawn at random out of box 1 and placed in box 2 without its color being observed.
A. If a ball is then drawn out of box 2, what is the probability that it is green?

I have equal probabilities of the ball drawn from box 1 is either R (1/2) or G (1/2)

if R is drawn from box 1 (Box 2 will now have RRRRGG)
p(G) 1/2* 1/3= 1/6

if G is drawn from box 1 (Box 2 will now have RRRGGG)
p(G) 1/2 * 1/2 = 1/4

A/ p(G) = 1/6 + 1/4 = 5/12

B. First, a box is chosen at random, and then a ball is drawn at random out of the chosen box. What is the probability that this ball is red?

p( box 1)= 1/2 * 2/4 = 2/8 = 1/4

p( box 2)= 1/4 * 3/5 = 3/10

p( red)= 1/4 + 3/10 = 11/20

Is this the correct way to calculate it? Thanks.

 

[AvgStudent]

Hi Zelda,
I like to do this with probability theories since it's more concrete for me. However, I got a different answer for A, but the same for B. Someone can double-check me.

Question A:
\(\displaystyle \Pr(D_2=G)=\Pr(D_2=G \cap D_1=G)+\Pr(D_2=G \cap D_1=R)\)
\(\displaystyle \Pr(D_2=G)= \Pr(D_2=G|D_1=G)\Pr(D_1=G) +\Pr(D_2=G|D_1=R)\Pr(D_1=R)\)
\(\displaystyle \Pr(D_2=G)= \frac{3}{6}\times \frac{1}{2} +\frac{4}{6}\times \frac{1}{2}=\boxed{\frac{7}{12}}\)

Question B, I used a similar idea:
\(\displaystyle \Pr(R)=\Pr(R \cap B_1)+\Pr(R \cap B_2)\)
\(\displaystyle \Pr(R)=\Pr(R |B_1)\Pr(B_1)+\Pr(R| B_2)\Pr(B_2)\)
\(\displaystyle \Pr(R)=\frac{2}{4}\times \frac{1}{2} + \frac{3}{5}\times \frac{1}{2}=\boxed{\frac{11}{20}}\)

Let me know if you have any questions. Hope this helps.

 

[Zelda22]

Hi Zelda,
I like to do this with probability theories since it's more concrete for me. However, I got a different answer for A, but the same for B. Someone can double-check me.

Question A:
\(\displaystyle \Pr(D_2=G)=\Pr(D_2=G \cap D_1=G)+\Pr(D_2=G \cap D_1=R)\)
\(\displaystyle \Pr(D_2=G)= \Pr(D_2=G|D_1=G)\Pr(D_1=G) +\Pr(D_2=G|D_1=R)\Pr(D_1=R)\)
\(\displaystyle \Pr(D_2=G)= \frac{3}{6}\times \frac{1}{2} +\frac{4}{6}\times \frac{1}{2}=\boxed{\frac{7}{12}}\)

Question B, I used a similar idea:
\(\displaystyle \Pr(R)=\Pr(R \cap B_1)+\Pr(R \cap B_2)\)
\(\displaystyle \Pr(R)=\Pr(R |B_1)\Pr(B_1)+\Pr(R| B_2)\Pr(B_2)\)
\(\displaystyle \Pr(R)=\frac{2}{4}\times \frac{1}{2} + \frac{3}{5}\times \frac{1}{2}=\boxed{\frac{11}{20}}\)

Let me know if you have any questions. Hope this helps.

why 3/5 ??? If you move a green, now you will have 6 balls total; 3R and 3G, I think should be 1/2 * 1/2 ????

 

[AvgStudent]

why 3/5 ??? If you move a green, now you will have 6 balls total; 3R and 3G, I think should be 1/2 * 1/2 ????

Hi Zelda,
Are you asking about A or B?

 

[BigBeachBanana]

if R is drawn from box 1 (Box 2 will now have RRRRGG)
p(G) 1/2* 1/3= 1/6

Your mistake is here. You have 4R and 6Balls, so the probability of drawing a red is 4/6=2/3, and not 1/3. The rest should match AvgStudent 's answer.

 

[Zelda22]

Hi Zelda,
Are you asking about A or B?

Found my error, it was a typo, when I reduced 3/6 = 1/2 I typed 1/3

if R is drawn from box 1 (Box 2 will now have RRRRGG)
p(G) 1/2* 1/2= 1/4

if G is drawn from box 1 (Box 2 will now have RRRGGG)
p(G) 1/2 * 2/3 = 1/3

A/ p(G) = 1/4 + 1/3= 7/12

 

[Zelda22]

Hi Zelda,
I like to do this with probability theories since it's more concrete for me. However, I got a different answer for A, but the same for B. Someone can double-check me.

Question A:
\(\displaystyle \Pr(D_2=G)=\Pr(D_2=G \cap D_1=G)+\Pr(D_2=G \cap D_1=R)\)
\(\displaystyle \Pr(D_2=G)= \Pr(D_2=G|D_1=G)\Pr(D_1=G) +\Pr(D_2=G|D_1=R)\Pr(D_1=R)\)
\(\displaystyle \Pr(D_2=G)= \frac{3}{6}\times \frac{1}{2} +\frac{4}{6}\times \frac{1}{2}=\boxed{\frac{7}{12}}\)

Question B, I used a similar idea:
\(\displaystyle \Pr(R)=\Pr(R \cap B_1)+\Pr(R \cap B_2)\)
\(\displaystyle \Pr(R)=\Pr(R |B_1)\Pr(B_1)+\Pr(R| B_2)\Pr(B_2)\)
\(\displaystyle \Pr(R)=\frac{2}{4}\times \frac{1}{2} + \frac{3}{5}\times \frac{1}{2}=\boxed{\frac{11}{20}}\)

Let me know if you have any questions. Hope this helps.

Thank you so much

 

[Zelda22]

Oh boy, now I get confused again.

BOX 2 (RRRGG)

If the ball moved to box2 is R, then box 2 will have (RRRRGG)
p(G) 1/2 * 2/6 = 1/6

***(I don't understand how you get the 4/6 ???

If the ball moved to box2 is G then box 2 will have (RRRGGG)
p(G) 1/2 * 3/6 = 1/4

p(G) = 1/6 + 1/4 = 5/12

 

[Cubist]

A typo in red (ought to be 1/2 not 1/4) but the = 3/10 is correct...

B. First, a box is chosen at random, and then a ball is drawn at random out of the chosen box. What is the probability that this ball is red?

p( box 1)= 1/2 * 2/4 = 2/8 = 1/4

p( box 2)= 1/ 4 * 3/5 = 3/10 <------------

p( red)= 1/4 + 3/10 = 11/20

 

[AvgStudent]

Oh boy, now I get confused again.

BOX 2 (RRRGG)

If the ball moved to box2 is R, then box 2 will have (RRRRGG)
p(G) 1/2 * 2/6 = 1/6

***(I don't understand how you get the 4/6 ???

If the ball moved to box2 is G then box 2 will have (RRRGGG)
p(G) 1/2 * 3/6 = 1/4

p(G) = 1/6 + 1/4 = 5/12

Hi Zelda,
You're correct that it should be 2/6. I was picking red, not green, so I got 4/6.

 

[Cubist]

You're correct that it should be 2/6. I was picking red, not green, so I got 4/6.

An unfortunate mistake because your post#2 is nicely presented This is what it would look like with a correction:-

Question A:
\(\displaystyle \Pr(D_2=G)=\Pr(D_2=G \cap D_1=G)+\Pr(D_2=G \cap D_1=R)\)
\(\displaystyle \Pr(D_2=G)= \Pr(D_2=G|D_1=G)\Pr(D_1=G) +\Pr(D_2=G|D_1=R)\Pr(D_1=R)\)
\(\displaystyle \Pr(D_2=G)= \frac{3}{6}\times \frac{1}{2} +\frac{\color{green}2\color{black}}{6}\times \frac{1}{2}=\boxed{\frac{\color{green}5\color{black}}{12}}\)

 

[Zelda22]

An unfortunate mistake because your post#2 is nicely presented This is what it would look like with a correction:-

Question A:
\(\displaystyle \Pr(D_2=G)=\Pr(D_2=G \cap D_1=G)+\Pr(D_2=G \cap D_1=R)\)
\(\displaystyle \Pr(D_2=G)= \Pr(D_2=G|D_1=G)\Pr(D_1=G) +\Pr(D_2=G|D_1=R)\Pr(D_1=R)\)
\(\displaystyle \Pr(D_2=G)= \frac{3}{6}\times \frac{1}{2} +\frac{\color{green}2\color{black}}{6}\times \frac{1}{2}=\boxed{\frac{\color{green}5\color{black}}{12}}\)

Yay! my originals answers were correct
5/12 and 11/20

 

[pka]

Box 1 (RRGG) Box 2 (RRRGG)
A ball is drawn at random out of box 1 and placed in box 2 without its color being observed.
A. If a ball is then drawn out of box 2, what is the probability that it is green?

The correct word is dependent: [imath]R_1[/imath] is the event that a red ball is drawn first likwise
[imath] G_1[/imath] for green. The question asks [imath]\mathcal{P}(G_2)[/imath] The events [imath]R_1~\&~G_1[/imath] partition the space.
The colour of second draw depends upon the colour of the first. That is one of two.
Thus [imath]\mathcal{P}(G_2)=\mathcal{P}(G_2\cap G_1)+{P}(G_2\cap R_1)[/imath] Next we use conditional probability to find intersection.
______________[imath]=\mathcal{P}(G_2| G_1)\mathcal{P}(G_1)+{P}(G_2|R_1)\mathcal{P}(R_1)[/imath] Recall that [imath]\mathcal{P}(A\cap B) =\mathcal{P}(B|A)\cdot\mathcal{P}(A)[/imath]
______________[imath]=\dfrac{3}{6}\cdot\dfrac{2}{4}+\dfrac{2}{6}\cdot\dfrac{2}{4}[/imath]
______________[imath]=\bf\dfrac{5}{12}[/imath][imath][/imath][imath][/imath][imath][/imath][imath][/imath][imath][/imath][imath][/imath]