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1) Two dice, each numbered 1 to 6, are tossed. Find the

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SVP
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1) Two dice, each numbered 1 to 6, are tossed. Find the [#permalink] New post 05 Oct 2005, 19:29
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1) Two dice, each numbered 1 to 6, are tossed. Find the probability that (a) at least one show a "4" (b) the total score is 9.

(2) If the dice are thrown n times, find the probability of obtaining (a) no double "4" (b) at least one double "4" in the n throws.
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 [#permalink] New post 05 Oct 2005, 20:08
Hi,
I would use the binomial theorem in all cases.

C(n,r)* p^k * p^(1-k), where k= prob of success.

Hope this helps.
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 [#permalink] New post 05 Oct 2005, 20:11
(1)
a) For the dice to have atleast one 4 = 1-(probability of no fours)

probability of 2 no fours = 5/6*5/6 = 25/36

1-(25/36)= 11/36

b) for total score to be 9, we have a few options (4,5),(5,4),(3,6) and (6,3), being the number rolled on each die respectively.

4/6*1/6= 4/36

(2)
Since each roll of the die is an independent event, I dont think the probability of a certain event occuring will change in 'n' rolls.

(a) probability of not rolling a double 4 = 1-probability of rolling a double 4.

probability of rolling a double 4 = 1/6*1/6=1/36. Probability of not rolling a double 4 = 1-(1/36)=35/36

(b) probability of rolling a double 4 = 1/6*1/6=1/36. This will not change in n tries, as each is an independent event.
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Re: Probability [#permalink] New post 06 Oct 2005, 01:00
HIMALAYA wrote:
1) Two dice, each numbered 1 to 6, are tossed. Find the probability that (a) at least one show a "4" (b) the total score is 9.

(2) If the dice are thrown n times, find the probability of obtaining (a) no double "4" (b) at least one double "4" in the n throws.


1a) = 11/36 (1 - p(of no 4 both times) = 1- (5/6*5/6)

1b) = 4/36 (There are 4 events that totals 9 of the 36 events so it is 4/36)

2a) = 35/36 ( 1 - p(of 4 both times) = 1- (1/6*1/6))

2b) = 1/36 (There is only 1 combination of 36 hence 1/36
Re: Probability   [#permalink] 06 Oct 2005, 01:00
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1) Two dice, each numbered 1 to 6, are tossed. Find the

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