Wednesday

April 16, 2014

April 16, 2014

Posted by **Sharon** on Wednesday, October 7, 2009 at 6:51am.

a.) No winning tickets?

b.) All winning tickets?

c.) At least one winning ticket?

d.) At least one non-winning ticket?

- probability -
**Reiny**, Wednesday, October 7, 2009 at 8:49amthere 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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