## Anita505 Group Title Two balls are drawn in succession out of a box containing 5 red and 5 white balls. Find the probability that at least 1 ball was red, given that the first ball was.. A) Replaced before the second draw B) NOT replaced before the second draw a) Find the probability that at least 1 ball was red, given that the first ball was replaced before the second draw. one year ago one year ago

1. goformit100 Group Title

I know the solution...

2. Anita505 Group Title

3. kropot72 Group Title

Section A) Consider the following 3 situations: (1) A: First ball red B: Second ball red The events A and B are independent therefore P(two red) = (5/10) * (5/10) ...........................(1) (2) A: First ball red B: Second ball black The events A and B are independent therefore P(one red) = (5/10) * (5/10) ...........................(2) (3) A: First ball black B: Second ball red The events A and B are independent therefore P(one red) = (5/10) * (5/10) ...........................(3) The probability that at least 1 ball was red, given that the first ball was replaced before the second draw is the sum of results (1), (2) and (3). Can you calculate?

4. Anita505 Group Title

Sorry i can not calculate it,

5. Anita505 Group Title

my calculator is currently not working

6. Anita505 Group Title

so this is what it would look like then (5/10)(5/10)+(5/10)(5/10)+(5/10)(5/10)=

7. kropot72 Group Title

You do not need a calculator to work this out: $P(at\ least\ one\ red)=(\frac{5}{10}\times \frac{5}{10})+(\frac{5}{10}\times \frac{5}{10})+(\frac{5}{10}\times \frac{5}{10})=?$

8. kropot72 Group Title

Hint: $\frac{5}{10}\times \frac{5}{10}=\frac{1}{2}\times \frac{1}{2}$

9. Anita505 Group Title

so the answer would be 0.75

10. kropot72 Group Title

11. Anita505 Group Title

oops no sorry my mistake

12. kropot72 Group Title

0.75 or 3/4 is correct.

13. Anita505 Group Title

How would i find the probability that at least 1 ball was red, given that the first ball s not replaced before the second draw?

14. teddyt477 Group Title

take the chance after by subtracting out already drawn balls

15. Anita505 Group Title

hmm?

16. kropot72 Group Title

Section B) Probability first ball is red = 5/10 Probability second ball is red = 4/9 P(two red) = (5/10 * 4/9) ................(1) Probability first ball is red = 5/10 Probability second ball is black = 5/9 P(one red) = (5/10 * 5/9) ...............(2) Probability first ball is black = 5/10 Probability second ball is red = 5/9 P(one red) = (5/10 * 5/9) ...............(3) $P(at\ least\ one\ red)=(\frac{1}{2}\times \frac{4}{9})+(\frac{1}{2}\times \frac{5}{9})+(\frac{1}{2}\times \frac{5}{9})=?$

17. Anita505 Group Title

thank you and the answer i have recieved was 0.77777777777

18. Anita505 Group Title

is that also what you got?

19. kropot72 Group Title

Your answer to B) is correct. Good work :)

20. Anita505 Group Title

thank you for your help! :)

21. Anita505 Group Title

but should i put my answer as 0.78?

22. Anita505 Group Title

it says to simply my answer into a integer or fraction

23. Anita505 Group Title

simplify**

24. Anita505 Group Title

oops nvm its 7/9 hahaha thanks!

25. kropot72 Group Title

To put the answer as a fraction just add the following fractions and simplify: $\frac{4}{18}+\frac{5}{18}+\frac{5}{18}=?$

26. kropot72 Group Title

Yes, 7/9 is correct!

27. Anita505 Group Title

:)

28. kropot72 Group Title

You're welcome :)