Tuesday

November 25, 2014

November 25, 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

- probability -
**keisha**, Saturday, November 8, 2014 at 7:57pmSixteen batteries are tested to see if they last as long as manufacturers claim.Four batteries fail the test. Two are selected at random without replacement. Find the probability that both batteries fail the test. Find the probability that both batteries pass the test. Find the probability that at least one battery fails the test. Which of the events can be considered unusual? Explain

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