Math homework concerning probability?

Question

I would like an explaination, not so much a answer, but an explaination that will help me to understand this problem.

1. Two urns each contain white balls and blacks balls. Urn 1 contains four white balls and two black balls and urn 2 contains six white balls and five black balls. A ball is drawn from each urn. What is the probalility that both balls are black?

Answer 1

The first urn holds a total of 6 balls, 2 of which are black. 2/6 = 1/3, so there is a 33% chance f the first ball being black.

The second urn has a total of 11 balls, 5 of which are black. 5/11, so there is about a 45% chance of the ball picked being black.

In order for both of them to be black, both conditions must occur. To find this you must multiply the two previous probabilites together to find the overall probability.


33% x 45% (1/3 x 5/11) which comes out to 5/33, or 15%.

Answer 2

Probability is the number of successful outcomes divided by the total outcomes.

The first urn has 2 black balls out of a total of 6 balls. So the chance that ball drawn from the first urn is black is 2/6 (or 1/3).

Similarly, the second urn has 5 black balls out of a total of 11 balls. Therefore, the chance that the ball drawn from the second urn is black is 5/11.

The chance that you have both events happening is the product:
1/3 * 5/11 = 5/33. This is because the first event happens 1/3 of the time and the second even happens 5/11 of those times. The probability is the product (5/33). If you turn it into a percentage,it is 15.1515...%

Answer 3

Simply explained, probability is the likelihood of an event occuring. So, if you roll a die, the probability of getting the number "1" on the die is 1 to 6--1 means there is only 1 side that says number "1." 6 means, there are a total of 6 sides on the die. In fractions, the probability becomes 1/6.

In your math homework, you will have two sets of probabilities. Probability number 1 comes from urn 1, and probability number 2 comes from urn 2. You will then have to add up the probabilities like you would add up fractions.

I will give you some of the answer to get you started. Urn 1 contains four white balls and two black balls. You have a total of 6 balls. Like the die example, you have 6 probabilities that may happen. The difference is, there is only one side that is painted number "1" on the die. In the urns, there are two black balls. So there is a greater probability of getting a black ball. The probability is 2 to 6, because there are 2 black balls among 6 balls total. In fractions, the probability is 2/6.

Apply the same concept to urn number 2, and then add up the fractions. You may have to cross-multiply to get the answer.

Good luck.

Answer 4

The outcome of the first selection has no affect on the outcome of the second selection since the balls are taken from different urns. Therefore we call these independent events.
For independent events, A and B, the probability of both A and B occurring is given by P(A)*P(B).
Consider the first urn: there are 2 black balls out of a total of 6 balls and so the probability of selecting black is 2 out of 6, ie
P(1st ball is black)=2/6=1/3
Similarly, for urn 2:
P(2nd ball is black) = 5/11
So P(both black) = 1/3 * 5/11 = 5/33

Hope that's clear!

Answer 5

in this case drawing the black ball from the urn is the sucess of the experiment. so when you have four white balls and two black balls in a urn your chance of drawing a black ball is 2/6, thats like saying you have a odd of 2 to 4 for the black ball. this is straight forward from the basic premise of probability, which is,

P(A)=(success of experiment)/(total sample space)

here the total sample space is 6(4 white and 2 black)

now for the second case, using the same logic, you have probability of 5/11 for the success of the second experiment.

since both the experiment are independent of each other i.e the probability of drawing a black ball from urn 1 does not affect the probability of drawing the black ball from urn 2, so the total probability is = (2/6)*(5/11).

note:i generally follow a simple rule if two experiments are INDEPENDENT and the question says "condition 1 AND condition 2", then multiply the individual probability "P(A)*P(B) , but it the question say "condition 1 OR condition 2" then add up the individual probability "P(A)+P(B)"

Answer 6

Here is my reasoning.

Prob(both balls drawn are black) = Prob(ball from Urn 1 is black)*Prob(Ball from Urn 2 is black) = [(number of black balls in Urn 1)/(number of balls in Urn 1)]*[(number of black balls in Urn 2)/(number of balls in Urn 2)] =

If you like you can take it from here or you can scroll down slightly for the answer.
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(2/6)*(5/11) = 10/66 = 5/33, or about 15%.

Answer 7

it id the prob that the first ball is black AND the 2nd ball is black.

1st black = good balls/ all balls = 2/6

2ndidem dito

since both should AND occur the final answer is the product of both