Probability

[TuanNguyen87]

The probability is 1/2 that a certain coin will turn up heads on any given toss. If the coin is to be tossed three times, what is the probability that on at least one of the tosses the coin will turn up tails?

A. 1/8
B. 1/2
C. 3/4
D. 7/8
E. 15/16

Thanks a lot everyone!

Answer: D

[cans]

prob(h)=1/2
thus prob(t)=1/2
p(atleast 1 tail in 3 tosses) = 1 - p(no tail in 3 tosses) = 1 - p(all heads in 3 tosses)
=1 - (1/2)(1/2)(1/2) = 1- 1/8 = 7/8
IMO D

[smackmartine]

+1 for D

[Ashley@VeritasPrep]

prob(h)=1/2
thus prob(t)=1/2
p(atleast 1 tail in 3 tosses) = 1 - p(no tail in 3 tosses) = 1 - p(all heads in 3 tosses)
=1 - (1/2)(1/2)(1/2) = 1- 1/8 = 7/8
IMO D

Just want to tack on here that the strategy Cans has wisely shown is a good strategy to use on ALL "at least" probability questions. "At least" will (in any problem) cover several different possible scenarios -- in this problem specifically, at least one tail covers the outcome of one tail, the outcome of two tails, and the outcome of three tails. But since the probabilities of all possible outcomes combined always add up to a total of 1 (by definition), it's safe -- and much easier -- in an "at least" problem simply to calculate the probability of NOT satisfying the condition, and since THAT probability plus the probability of satisfying the condition must sum to 1, the probability of satisfying the condition simply equals 1 - P(not satisfying). The only way NOT to satisfy the condition given in this problem of "at least one tail" is to get ALL heads, and out of our eight possible outcomes, only one features all heads. Therefore P(all heads) = 1/8, and from there, as Cans says, P(at least one tail) = 1 - 1/8 = 7/8.

One more related trick. This one's not applicable to this particular problem (or to "at least" problems in general), but it's another instance of its being wise to shift to thinking about heads when a problem asks about tails, or vice versa. This is the "all but one" probability problem. Suppose we are flipping five pennies in a row, and the problem asks us for the probability of producing exactly four tails. This is then an instance in which the condition is that ALL BUT ONE coin come up tails -- so in other words, to satisfy this condition means, equivalently, flipping EXACTLY ONE head. It's much easier to compute how many ways there are to get exactly one head (five ways, since the head could show up on any single one of the five pennies) than it is to think about the problem in terms of how many ways there are to get four tails. Since getting exactly one head is actually the exact SAME scenario as getting exctly four tails, we *don't* do the "1 minus" business -- we just wind up straight at our answer (in this case [spoiler]5/32[/spoiler]).

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There is one more method for all the coin toss questions, a little lengthy but very accurate...

Let's say we first calculate the Number of possible outcomes. This is very important in probability sums.

N = {HHH ; HHT ; HTH ; THH ; TTH ; THT ; HTT ; TTT}

You can see there are 8 possible outcomes when the coin is tossed 3 times. H = Heads
T = Tails
so we want atleast one tail in any of the times the coin being tossed.

S ={HHT ; HTH ; THH ; TTH ; THT ; HTT ; TTT}

This is the desired result. So the overall answer is (7/8).

I really hope this calculation helps understanding....