someth1ng
Retired Nov '14
I just was looking at random MX2 questions and I came across this, it was on a trial paper in Q7 - I could be totally wrong since I haven't done any 4U probability...
A box contains 6 cards, two of them are identical. From this box 3 cards are drawn without replacement.
(i) How many different selections could be made.
(ii) What is the probability that a selection will include two identical cards.
My 3U style solutions:
(i) You can pick none, one or two of the identical cards so:
Two I (I, I, 4 options) = 4C1 = 4 ways
One I (I, 4 options, 3 options) = 4C2 = 6 ways
No I (4 options, 3 options, 2 options) = 4C3 = 4 ways
Total = 4+6+4 = 14 ways
(ii) P(Two I) = 4/14 = 2/7
Am I right?
A box contains 6 cards, two of them are identical. From this box 3 cards are drawn without replacement.
(i) How many different selections could be made.
(ii) What is the probability that a selection will include two identical cards.
My 3U style solutions:
(i) You can pick none, one or two of the identical cards so:
Two I (I, I, 4 options) = 4C1 = 4 ways
One I (I, 4 options, 3 options) = 4C2 = 6 ways
No I (4 options, 3 options, 2 options) = 4C3 = 4 ways
Total = 4+6+4 = 14 ways
(ii) P(Two I) = 4/14 = 2/7
Am I right?
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