## anonymous 3 years ago 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.

1. goformit100

I know the solution...

2. anonymous

3. kropot72

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. anonymous

Sorry i can not calculate it,

5. anonymous

my calculator is currently not working

6. anonymous

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

7. kropot72

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

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

9. anonymous

so the answer would be 0.75

10. kropot72

11. anonymous

oops no sorry my mistake

12. kropot72

0.75 or 3/4 is correct.

13. anonymous

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. anonymous

take the chance after by subtracting out already drawn balls

15. anonymous

hmm?

16. kropot72

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. anonymous

thank you and the answer i have recieved was 0.77777777777

18. anonymous

is that also what you got?

19. kropot72

20. anonymous

thank you for your help! :)

21. anonymous

but should i put my answer as 0.78?

22. anonymous

it says to simply my answer into a integer or fraction

23. anonymous

simplify**

24. anonymous

oops nvm its 7/9 hahaha thanks!

25. kropot72

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

Yes, 7/9 is correct!

27. anonymous

:)

28. kropot72

You're welcome :)