Posted by Sharon on Wednesday, October 7, 2009 at 6:51am.
there are 346 non-winning tickets in the batch.
So the prob of all 4 being non-winning is
(346/350)(345/349)(344/348)(343/347)
= .9549
or, if you know the C(n,r) combination topic,
it would be C(346,4)/C(350,4) = .9549
b) prob = (4/350)(3/349)(2/348)(1/347)
= .000000001
or C(4,4)/C(350,4) = .000000001
c) at least one winning ticket
= 1 - prob(all losing tickets)
= 1 - .9549 = .0451
d) at least one non-winner
= 1 - .000000001 = .999999999
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