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A box contains 6 blue balls, 4 black balls and 5 red balls at the same size . A ball is selected at random from the box and then replaced . A second ball is then selected. Find the probability of obtaining ; (i) two red balls ; (ii) two blue balls or two black balls ; (iii) one black ball and one red balls.​

User Zantier
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1 Answer

3 votes

Answer:

(i) 1/21

(ii) 1/10

(iii) Take a look at the explanation: Try this one yourself. I have given you some hints.

Explanation:

(i) Two red balls:

To do this, we need to find the total amount of possible choices first. To do this, we multiply 15 by 14. This is our denominator:

15(14) = 210

Now, we need to find the total combinations of red balls. We solve 5 choose 2 for this one.

5 choose 2 = 5(4)/2! = 10

Our numerator is 10. Therefore, our probability is 10/210 = 1/21.

(ii) Two blue balls or two black balls:

To do this, we need to add the probabilities of getting a blue ball with a black ball. (Since there is an "or" sitting there. If there is an "and", we multiply)

So, let's calculate the probability of getting a blue ball first:

Blue:

We use the same denominator as before: 210.

Our numerator is now 6 choose 2, which is:

6 choose 2 = 6(5)/2! = 15.

Now, our fraction is 15/210, BUT, dont simplify, as we will need to add.

Black:

Same steps: denominator is 210, but the numerator is 4 choose 2.

Solving 4 choose 2:

4 choose 2 = 4(3)/2! = 6.

Our numberator is 6.

But, we cant forget to add them!

(15 + 6)/210 is 21/210, which is 1/10.

(iii) I'll let you try this one by yourself. Here is a hint:

Solve for the probability of chooseing a black ball and a red ball. Then, multiply.

Enjoy the process, and I hoped this helped you! (If you have any questions, feel free to ask)

User Mike Debela
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