http://www.xtremepapers.net/CIE/International A And AS Level/9709 - Mathematics/9709_s07_qp_6.pdf
question 7 last two parts
question 7 last two parts
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paper 62
- Thread starter ahmed t
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You don't have to care the MS's answer as long as yours is correct and REASONABLE.
By now I have thought of two ways to solve part (ii).
Method 1: Direct Probability Addition:
The three including 2 green can be: GG-, G-G and -GG
So calculate the P for each case, then add together.
GG- = G-G = -GG = (4/12)(3/11)(8/10) = 4/55
So total P = 3 * (4/55) = 12/55
Method 2: Combination dividing
The number of ways to choose 2 green is 4C2 * 8C1 = 48
Number of ways to choose ANY 3 with no restrictions is 12C3 = 220
The P is then 48/220 = 12/55
By now I have thought of two ways to solve part (ii).
Method 1: Direct Probability Addition:
The three including 2 green can be: GG-, G-G and -GG
So calculate the P for each case, then add together.
GG- = G-G = -GG = (4/12)(3/11)(8/10) = 4/55
So total P = 3 * (4/55) = 12/55
Method 2: Combination dividing
The number of ways to choose 2 green is 4C2 * 8C1 = 48
Number of ways to choose ANY 3 with no restrictions is 12C3 = 220
The P is then 48/220 = 12/55
- Messages
- 260
- Reaction score
- 104
- Points
- 53
Often about 45 out of 50... Like in the recent mock exams.