Please help me with this study guide as soon as possible. I am need of immediate help since I have a test tomorrow. 

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jerichorayel's profile pic

jerichorayel | College Teacher | (Level 2) Senior Educator

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10 balls: 2 white, 3 red and 5 yellow.

a.i. 2 ways out of 10. 2/10 = 1/5

a.ii. 5 ways out of 10. 5/10 = 1/2

c.i. The three balls can't be white since there are only two white balls in the group. We will pick 3 red balls out of 3 available red balls  (3C3) and pick 3 yellow balls out of 5 available yellow balls (5C3).

The expression would be: (3C3 + 5C3)/10C3

c.ii. Picking one ball per color is represented as 2C1, 3C1 and 5C1. The expression would be: (2C1*3C1*5C1)/10C3

Sources:
rimmery's profile pic

rimmery | Student, Undergraduate | (Level 1) Honors

Posted on

a.i) There are 2 white balls in a total of 10 possible balls drawn. probability is 2/10 or 1/5=0.2

a.ii) Similarly, there's 5 yellow in a total of 10, so 5/10=1/2=0.5

b.i) There's 3 cases: when both balls are white, when they're both red, and when they're both yellow. The probability that two balls are the same colour is the sum of these three probabilities because they're separate outcomes.

White: From 2 white balls, you choose 2 of them, so you have (2C2). 

Then from the total of 10 balls, you choose 2 randomly, so the TOTAL number of possible 2-ball outcomes are (10C2).

The probability two balls are both white is (2C2)/(10C2)

Red: 3 red balls, choose 2. (3C2). Probability is (3C2)/(10C2)

White: (5C2)/(10C2)

Total = [(2C2)+(3C2)+(5C2)]/(10C2)

b.ii) Calculated in the first part, the probability is (3C2)/(10C2)

b.iii) This is a bit different since they didn't specify what the first colour can be. So if the first one is red, then out of 2 red balls left, you have to choose one of them (2C1), and the total number of possible ways to choose one more ball out of the 9 left is (9C1).  so you get (2C1)/(9C1). (Note that you do not care about what colour specifically the first ball is, so it is completely left out of the equation)

However if the first ball is not red, then you have a 3/9 chance of selecting a red ball since there are 3 red balls, and there are 9 balls left to choose from.

Another way to look at this is (3C1)/(9C1) = 3/0 = 1/3

The total is then (2C1)/(9C1)+1/3=5/9

c.i) So only two cases are possible this time. 

Red: 3 red balls in total, you have to choose 3. (3C3)

Yellow: 5 in total, choose 3 of them, (5C3)

There are 10 balls in total to choose 3 balls from, so

Probability=(3C3+5C3)\(10C3) 

c.ii) This is a bit complicated with the wording, from how I understand, it means that all three cannot be the same colour, but two can be the same. so what you do is for each of the cases in b.i, you add to the denominator the probability of other colours.

For example, If two balls are white, then the 3rd one is either red or yellow, and so there is a 100% chance of that happening since there is only 2 white balls.

If two balls are red, then from the 2white+5yellow balls left, you choose one of them. The probability of that is (7C1)/(8C1) Since there is 8 balls left in total.

Similarly for 2 yellow balls.

c.iii) If the third is the only red, then you have 7 balls left, and 3 red ones to choose from, so (3C1)/(7C1)

Phew, this has got to be my longest answer yet lol...

rachellopez's profile pic

rachellopez | Student, Grade 12 | (Level 1) Valedictorian

Posted on

When setting up a problem like this, the probability would equal the number of colored balls over the total number of balls. 

a.i. There are 2 white balls out of a total of 10 balls. 2/10 simplifies to 1/5.

a.ii. There are 5 yellow balls out of 10 total. 5/10 simplifies to 1/2.

b.ii. There are 3 red balls out of 10, so that would be 3/10 for the probability of the first red ball. As for the second, there would be only 2 red balls left and only 9 total. This fraction would be 2/9.

b.iii. For this, lets assume the first ball drawn is not red. After the first ball is drawn there are only 9 total and out of those there would be 3 that were red. That is 3/9 or 1/3.

Sorry I couldn't help with the rest.

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