Question

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.

Answer 1

I know the solution...

Answer 2

Can you please help me?

Answer 3

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?

Answer 4

Sorry i can not calculate it,

Answer 5

my calculator is currently not working

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

so the answer would be 0.75

Answer 10

Good work! Your answer is correct.

Answer 11

oops no sorry my mistake

Answer 12

0.75 or 3/4 is correct.

Answer 13

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?

Answer 14

take the chance after by subtracting out already drawn balls

Answer 15

hmm?

Answer 16

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)
\[P(at\ least\ one\ red)=(\frac{1}{2}\times \frac{4}{9})+(\frac{1}{2}\times \frac{5}{9})+(\frac{1}{2}\times \frac{5}{9})=?\]

Answer 17

thank you and the answer i have recieved was 0.77777777777

Answer 18

is that also what you got?

Answer 19

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

Answer 20

thank you for your help! :)

Answer 21

but should i put my answer as 0.78?

Answer 22

it says to simply my answer into a integer or fraction

Answer 23

simplify**

Answer 24

oops nvm its 7/9 hahaha thanks!

Answer 25

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

Answer 26

Yes, 7/9 is correct!

Answer 27

:)

Answer 28

You're welcome :)