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coins tossed

by shibal » Sat Jun 27, 2009 7:08 am
what is the difference (approach) between these two exercises?

Five fair coins are tossed. What is the probability that exactly three of the coins land tails side up?

Five fair coins are tossed. What is the probability that at least three of the coins land tails side up?
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by Robinmrtha » Sat Jun 27, 2009 7:32 am
Five fair coins are tossed. What is the probability that exactly three of the coins land tails side up?

Solution

Exactly three of the coins land tails up = probability of 1st coin landing tails
and probability of 2nd coin landing tails and probability of 3rd coin landing tails
=1/2 x 1/2 x1/2
=1/8

Five fair coins are tossed. What is the probability that at least three of the coins land tails side up?

Solution
probability that at least three of the coins land tails side up =Probability of three coins landing tails up or Probability of 4 coins landing tails up or Probability of 5 coins landing tails up

=1/8 +1/16 +1/32
=7/32
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by shibal » Sat Jun 27, 2009 7:37 am
Robinmrtha wrote:Five fair coins are tossed. What is the probability that exactly three of the coins land tails side up?

Solution

Exactly three of the coins land tails up = probability of 1st coin landing tails
and probability of 2nd coin landing tails and probability of 3rd coin landing tails
=1/2 x 1/2 x1/2
=1/8
i've got 5/16....

possible outcomes 32 (2^5); different ways to have 3 tails 5!/3!(5-3)! = 10
10/32 --> 5/16...
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by DanaJ » Sat Jun 27, 2009 8:35 am
@shibal, you make a pretty good point. While the first answer is not entirely wrong, it only considers the situation where you have exactly the first three cases as your tails.

The way you've presented it makes sense, since it also takes into consideration the option that the three tails appear in other line-ups.

For your second question, the answer will indeed be probability of 3 + probability of 4 + probability of 5. You've already calculated the probability for 3 (5/16). The probability of 4 will be 5/32 or 1/8, since you have five available positions for that stray heads in there:

T T T T H
T T T H T
T T H T T
T H T T T
H T T T T

Getting all tails will obviously be 1/32. Add them all up to get 1/2. This also makes sense if you look at it this way: the probability you're looking for is 1 - (0, 1, 2 tails) which is 1 - (5, 4, 3 heads) or 1 - (the probability of getting at least three heads). Since the probability of getting heads or tails is the same, the probabilities for at least three heads or tails will be exactly the same.
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