SOLUTION: https://vle.mathswatch.co.uk/images/questions/question16415.png please help

Question 1118089: https://vle.mathswatch.co.uk/images/questions/question16415.png
please help
Answer by math_helper(2461) (Show Source): You can put this solution on YOUR website!
Let's look at the four cases:
Let P(R,R) = probability of selecting red then another red
What value is P(R,R)?
On the first pick, there are 7 red and 11 total, so P(R) = 7/11
On the 2nd pick, there are 6 red and 10 total, so P(R,R) = (7/11)*(6/10) = 42/110
1. P(R,R) = 42/110

Similarly, using W for white:
2. P(R,W) = (7/11)*(4/10) = 28/110
3. P(W,R) = (4/11)*(7/10) = 28/110
4. P(W,W) = (4/11)*(3/10) = 12/110

Notice these cover all possible outcomes, as expected. We know this because the numerators, when added, sum to 110 and 110/110=1

The probability of one of each color is just the sum of case 2, P(R,W), and case 3, P(W,R):
28/110 + 28/110 = or about 0.51
I hope this helps you.

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