# probablility with balls

• May 30th 2010, 07:07 AM
andyboy179
probablility with balls
Hi, this is my question:

Harry has a bag of different coloured balls, 3 red balls, 2 yellow balls and 1 blue ball, identical except for their colour.
Harry picks two balls from the bag at random and without replacement.

A) What is the probability that Harry picks at least one yellow ball?

for this question would i do, 2/6 x 1/5= 2/30= 1/15 or would it be 2/6 + 1/5= 3/11?? Or is that wrong?
Could you tell me which one it is or if its not one of them tell me where im going wrong and how i can improve it!
Thanks alot!!
• May 30th 2010, 07:41 AM
e^(i*pi)
Quote:

Originally Posted by andyboy179
Hi, this is my question:

Harry has a bag of different coloured balls, 3 red balls, 2 yellow balls and 1 blue ball, identical except for their colour.
Harry picks two balls from the bag at random and without replacement.

A) What is the probability that Harry picks at least one yellow ball?

for this question would i do, 2/6 x 1/5= 2/30= 1/15 or would it be 2/6 + 1/5= 3/11?? Or is that wrong?
Could you tell me which one it is or if its not one of them tell me where im going wrong and how i can improve it!
Thanks alot!!

It is easier to find the chance that Harry does not pick a yellow ball and subtract it from 1.

There are 4 non-yellow and 2 yellow. Hence there is a $\frac{4}{6}$ that he does not pick a yellow ball first time around.

Second time there are now 5 balls, 3 of which are not yellow which means there is a $\frac{3}{5}$ he does not pick a yellow ball.

Because this is an and scenario we multiply: $\text{P'(Y)} = \frac{4}{6} \times \frac{3}{5} = \frac{2}{5}$

Subtract this from 1 to find the chance of at least one yellow
• May 30th 2010, 07:41 AM
Rapha
Hi!

Quote:

Originally Posted by andyboy179
Hi, this is my question:

Harry has a bag of different coloured balls, 3 red balls, 2 yellow balls and 1 blue ball, identical except for their colour.
Harry picks two balls from the bag at random and without replacement.

A) What is the probability that Harry picks at least one yellow ball?

for this question would i do, 2/6 x 1/5= 2/30= 1/15 or would it be 2/6 + 1/5= 3/11?? Or is that wrong?
Could you tell me which one it is or if its not one of them tell me where im going wrong and how i can improve it!
Thanks alot!!

There are 6 balls (2 of them are yellow)

so P("picking a yellow ball with the first try") = 2/6
P("not picking a yellow ball with the first try") = 4/6

A) We need to calculate

P("picking a yellow ball with the first try and picking a non-yellow-ball with the second try")

This is equal to 2/6 * 4/5 (this is because of "without replacement" after picking a ball, there are 5 remaining)

Anyway there are more events:

Most obvious one: picking a non-yellow ball and then a yellow ball

P((non-yellow, yellow)) = 4/6*2/5

And another possible event:

P("Picking a yellow ball with the first try and picking a yellow ball with the second try")

After picking a yellow ball (probability equals 2/6 ) there are 5 balls remaining, one of them is yellow
picking another yellow ball -> 1/5

So
P("Picking a yellow ball with the first try and picking a yellow ball with the second try") = P(yellow, yellow ) = 2/6 *1/5

And therefor the solution is

P(yellow, non-yellow) + P(non-yellow, yellow) + P(yellow, yellow)

Any questions?

Rapha

Edit: This is a more complicated way. It is better/easier to solve it like e^(i*pi)'s.

He calculated

1- P(non-yellow, non-yellow) = P(at least one yellow ball)

The solution should be the same.
• May 30th 2010, 07:47 AM
Quote:

Originally Posted by andyboy179
Hi, this is my question:

Harry has a bag of different coloured balls, 3 red balls, 2 yellow balls and 1 blue ball, identical except for their colour.
Harry picks two balls from the bag at random and without replacement.

A) What is the probability that Harry picks at least one yellow ball?

for this question would i do, 2/6 x 1/5= 2/30= 1/15 or would it be 2/6 + 1/5= 3/11?? Or is that wrong?
Could you tell me which one it is or if its not one of them tell me where im going wrong and how i can improve it!
Thanks alot!!

Hi andyboy179,

you first calculated the probability that both are yellow.
Yes you multiply them because these are stages of an event, the event of getting 2 yellows,
not different ways to get a successful outcome, which is at least one yellow.

There is also the chance that he will pick one yellow only.
The yellow could be picked first, followed by a non-yellow.
Or the yellow could follow the non-yellow.

You must calculate those 3 probabilities independently and then sum them.

Alternatively, you could calculate the probability of neither being yellow.
Then subtract that answer from 1, since all the other pairs will have at least one yellow and all the probabilities in a given situation sum to 1.
• May 30th 2010, 07:49 AM
Sudharaka
Quote:

Originally Posted by andyboy179
Hi, this is my question:

Harry has a bag of different coloured balls, 3 red balls, 2 yellow balls and 1 blue ball, identical except for their colour.
Harry picks two balls from the bag at random and without replacement.

A) What is the probability that Harry picks at least one yellow ball?

for this question would i do, 2/6 x 1/5= 2/30= 1/15 or would it be 2/6 + 1/5= 3/11?? Or is that wrong?
Could you tell me which one it is or if its not one of them tell me where im going wrong and how i can improve it!
Thanks alot!!

Dear andyboy179,

Although e^(i*pi)'s approch seem much more easier, you can also use this method.

First we can define the events as,

Event that he will get a yellow ball in the first turn= $R_{1}$

Event that he will get a yellow ball in the second turn= $R_{2}$

Therefore,

The probability that he will pick at least one yellow ball=(picking a yellow ball in the first turn and another ball(which is not yellow) in the second turn) or (pick a ball(not yellow) in the first turn and pick a yellow ball in the second turn) or (picking two yellow balls)

The probability that he will pick at least one yellow ball= $P[(R_{1}\cap{R'_{2}})\cup{(R'_{1}\cap{R_{2}})}\cup{( R_{1}\cap{R_{2}})}]$

Since the events, $(R_{1}\cap{R'_{2}})$, $(R'_{1}\cap{R_{2}})$ and $(R_{1}\cap{R_{2}})$ are mutually exclusive,

The probability that he will pick at least one yellow ball= $P(R_{1}\cap{R'_{2}})+{P(R'_{1}\cap{R_{2}})}+{P(R_{ 1}\cap{R_{2}})}$

Also the events $R_{1}, R'_{1}, R_{2}~and~R'_{2}$ are independent.

Therefore,

The probability that he will pick at least one yellow ball= $P(R_{1}).P(R'_{2})+P(R'_{1}).P(R_{2})+P(R_{1}).P(R _{2})$

The probability that he will pick at least one yellow ball= $\left(\frac{2}{6}\times\frac{1}{5}\right)+\left(\f rac{4}{6}\times\frac{2}{5}\right)+\left(\frac{2}{6 }\times\frac{1}{5}\right)=\frac{2}{30}+\frac{8}{30 }+\frac{2}{30}=0.4$

• May 30th 2010, 12:00 PM
andyboy179
i still a little confused but could i use a tree diagram and do this to work it out? (Y=yellow NY= not yellow)

Attachment 17053

then do the calculations?
• May 30th 2010, 12:15 PM
andyboy179
Quote:

Originally Posted by e^(i*pi)
It is easier to find the chance that Harry does not pick a yellow ball and subtract it from 1.

There are 4 non-yellow and 2 yellow. Hence there is a $\frac{4}{6}$ that he does not pick a yellow ball first time around.

Second time there are now 5 balls, 3 of which are not yellow which means there is a $\frac{3}{5}$ he does not pick a yellow ball.

Because this is an and scenario we multiply: $\text{P'(Y)} = \frac{4}{6} \times \frac{3}{5} = \frac{2}{5}$

Subtract this from 1 to find the chance of at least one yellow

so would the answer be 1/5 because you subtract 1 from 2/5?
• May 30th 2010, 12:25 PM
e^(i*pi)
Quote:

Originally Posted by andyboy179
so would the answer be 1/5 because you subtract 1 from 2/5?

$P(Y) = 1-\frac{2}{5}$

Note that $1 = \frac{5}{5}$
• May 30th 2010, 12:27 PM
andyboy179
Quote:

Originally Posted by e^(i*pi)
$P(Y) = 1-\frac{2}{5}$

Note that $1 = \frac{5}{5}$

thanks, could i use a tree diagram to get this answer?
• May 30th 2010, 12:36 PM
Quote:

Originally Posted by andyboy179
i still a little confused but could i use a tree diagram and do this to work it out? (Y=yellow NY= not yellow)

Attachment 17053

then do the calculations?

Yes!

that will work fine,
try to work out the probabilities that contain one yellow or two yellows in two consecutive branches.
Check whether you should be multiplying or adding probabilities at each branch.

Sum them all up at the end.

The alternative is....

find the consecutive branches with no yellow.
Calculate the probability of not getting a yellow.
All probabilities must sum to 1, in other words, the 4 final probabilities in this case.

Subtract the probability for no yellows from 1 for the probability of at least 1 yellow.
You can subtract fractions if you write both with a common denominator.
Then you can subtract numerators.

$1=\frac{2}{2}=\frac{3}{3}=\frac{4}{4}=\frac{5}{5}= ....$

Or convert your fraction to decimal before subtracting it from 1.
• May 30th 2010, 12:42 PM
andyboy179
would this work:
2/6 x 1/5 = 2/30= 1/15
2/6 x 4/5= 8/30= 4/15
4/6 x 2/5= 8/30= 4/15

1/15 + 4/15 + 4/15= 9/15?
• May 30th 2010, 12:50 PM
Quote:

Originally Posted by andyboy179
would this work:
2/6 x 1/5 = 2/30= 1/15 probability both are yellow
2/6 x 4/5= 8/30= 4/15 probability 1st is yellow, 2nd isn't
4/6 x 2/5= 8/30= 4/15 probability 1st is not yellow, 2nd is

1/15 + 4/15 + 4/15= 9/15? sum of the seperate probabilities for the sets of branches

Yes!(Clapping)
• May 30th 2010, 12:54 PM
e^(i*pi)
Quote:

Originally Posted by andyboy179
would this work:
2/6 x 1/5 = 2/30= 1/15
2/6 x 4/5= 8/30= 4/15
4/6 x 2/5= 8/30= 4/15

1/15 + 4/15 + 4/15= 9/15?

Yes, but you should cancel a factor of 3 from 9/15
• May 30th 2010, 12:57 PM
andyboy179
thanks alot! i think im starting to understand probability a bit more now! Also thanks to everone else that posted a n answer to my question(Clapping)(Clapping)(Clapping)