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

Although Bunuel's approach is, also in my opinion, the best way to go for this sort of question, you could also arrive at the same answer by using the following line of thought:

To get at least 1 tails, you can get one of the 3 configurations (in no particular order):

H H T -> 3 * 1/2*1/2*1/2
H T T -> 3 * 1/2*1/2*1/2
T T T -> 1/2*1/2*1/2

P = 3/8 + 3/8 + 1/8 = 7/8

The point is that two heads and a tail can occur in three ways: HHT, HTH, THH. The probability of each case is 1/2*1/2*1/2.


_________________

Yes, that's correct, the power must be the number of tries.

For the original question: P(at last 1 tails) = 1 - P(all heads) = 1 - (1/2*1/2*1/2)= 1 - (1/2)^3 = 7/8.

For your example: P(at least 1 blue) = 1 - P(no blue) = 1- (2/3*2/3*2/3*2/3) = 1- (2/3)^4.

Hope it's clear.
_________________

First thing to do is to come up with the total number of possible outcomes:
The coin is tossed 3 times and there is an equal probability that the coin will turn up heads or tail on each toss (which means that each toss has only two possible outcomes)
_ _ _
2*2*2=2^3=8
Multiply the number of possible outcomes per toss to arrive at the total number of possible outcomes. If the Question would state that the coin is to be tossed four times, the total number of possible outcomes would simply imply another multiplication by 2 or 2^4 which is 16.

Next step is to find the number of scenarios that fulfill the condition the Question stem asks for. (At least one tail)
One can easily recognize that ALL scenarios BUT ONE will include at least one tail. I am talking about the scenario in which all three tosses result in heads.
--> HHH
So 7 out of 8 scenarios will include at least one tail. THH, HTH, HHT etc…
This is already your final answer. P(at least one tail) = 7/8

Added info : Formula to solve similar questions

Let 'p' be the probability of getting a head and 'q' be the probability of getting a tail.
If 'n' coins are tossed or one coin is tossed 'n' times

Then \(^nC_r * p ^ r * q^{n-r}\) gives the probability of having r heads and n-r tails.

Hey dave13

RULE: The probability of an event occurring can never be greater than 1.

So, whenever during the course of a problem you get a number
greater than 1 for a probability, be assured that you have made a mistake!

You need to multiply the probability when the events are to occur simultaneously, or consecutively. 
For example, when we get three heads,
the probability will be \(\frac{1}{2}*\frac{1}{2}*\frac{1}{2} = \frac{1}{8}\).

We need to add probabilities when the events are alternatives.
For example, when there are three cases that can happen.
You can get HHT, HTT, TTT(for the three coins problem) as possible options for at least one tail.
In order to get the final answer, you need to add the individual probabilities for all the three cases.

Coming back to the problem
An easy way to solve this question is to find the probability that no tail
turns up and reduce that from 1(the maximum probability that an event can occur)

The probability that a tail does not turn up when three coins are tossed is when we get a head
on each of these coin tosses, which is \(\frac{1}{2}*\frac{1}{2}*\frac{1}{2} = \frac{1}{8}\)

Probability (atleast one tail turns up) = 1-P(All heads) = \(1 - \frac{1}{8} = \frac{7}{8}\)(Option D)

Hope this helps you!
_________________

When it comes to probability questions involving "at least," it's best to try using the complement.

That is, P(Event A happening) = 1 - P(Event A not happening)

So, here we get: P(getting at least 1 tails) = 1 - P(not getting at least 1 tails)
What does it mean to not get at least 1 tails? It means getting zero tails.
So, we can write: P(getting at least 1 tails) = 1 - P(getting zero tails)

Now let's calculate P(getting zero tails)
What needs to happen in order to get zero tails?
Well, we need heads on the first toss and heads on the second toss and heads on the third toss.
We can write P(getting zero tails) = P(heads on 1st AND heads on 2nd AND heads on 3rd)
This means that P(getting zero tails) = P(heads on 1st) x P(heads on 2nd) x P(heads on 3rd)
Which means P(getting zero tails) = (1/2)x(1/2)x(1/2)= 1/8

We're now ready to answer the question.
P(getting at least 1 tails) = 1 - P(not getting at least 1 tails)
= 1 - 1/8
= 7/8

Answer: D

Cheers,
Brent
_________________