The probability is 1/2 that a certain coin will turn up head

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

Hi,

Kindly explain why it is 3* 1/2 * 1/2 * 1/2.
According to my understanding, the prob of head is 1/2 and tail is 1/2.
So HHT just has to be 1/2 * 1/2 * 1/2.. Isn't it? why multiply by 3?
Kindly clarify.

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.

Bunuel, would you mind explaining how you find the power to which you have to raise? What if e.g. there is a bag with three marbles, blue, red and yellow. Now the question is e.g. "What is the probabilty to get a blue marble on at least 1 try if you try 4 times" (putting the marbles back all the time).

Would it be P(at least 1 blue marble) = 1 - P(none blue) = 1 - (2/3)^4 ??

Thanks for your explanation!

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.

ok great thank you!

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

In this problem, there are only two events that could occur for the 3 coin flips. Either the coin will land on tails zero times, or the coin will land on tails at least one time. (Remember that the phrase "at least one time" means "one or more."

Writing this in a probability statement yields:

P(landing on tails at least 1 time) + P(landing on tails zero times) = 1

Thus, we can say:

P(landing on tails at least 1 time) = 1 - P(landing on tails zero times)


Since we are tossing the coin 3 times, the outcome of zero tails in 3 tosses is the same as getting heads on all 3 tosses. We can calculate the probability of zero tails in 3 tosses as the probability of 3 heads in 3 tosses:

½ x ½ x ½ =1/8

Plugging this into our formula we have:

P(landing on tails at least 1 time) = 1 – 1/8

P(landing on tails at least 1 time) = 7/8

Answer is D.

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.

let Tails be T and Heads be H so i need to toss ONE coin three times and the probability is as follows:

first attempt and probability
T
H
H

second attempt and probability
H
T
H

third attempt and probability
H
H
T

so what am i to do next ? after i tossed coin 3 times.... Why do you guys above me all mutiplying ?

we have one coin and it does not effect the next event so we need to add up, no ? hmm like this 1/2+1/2+1/2 =3/2 i thought this was answer...

When do i need to multiply or add in probabality ? Difference?

help niks18

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!

Hello Bunuel

if two heads and a tail can occur in three ways: HHT, HTH, THH. The probability of each case is 3* 1/2*1/2*1/2.

then two tails and one heads can occure 3 times as well TTH, THT, HTT right ? 3 * 1/2*1/2*1/2.

and probabilty that all tails and zero heads TTT 1/2*1/2*1/2 (so here we dont need to multiply by 3 because there is only one combination)

am i thinking correctly ? thanks

Given that a coin is tossed three times and we need to find what is the probability that on at least one of the tosses the coin will turn up tails?

As the coin is tossed three times then number of cases = \(2^3\) = 8

P(At least one Tail) = 1 - P( 0 Tail) = 1 - P(HHH)

Out of the 8 cases there is only one case in which we get three Heads (HHH)

=> P(HHH) = \(\frac{1}{8}\)

=> P(At least one Tail) = 1 - P(HHH) = 1 - \(\frac{1}{8}\) = \(\frac{8 - 1}{8}\) = \(\frac{7}{8}\)

So, Answer will be D
Hope it helps!

Watch the following video to learn How to Solve Probability with Coin Toss Problems

To find the probability that on at least one of the three coin tosses the coin will turn up tails, we can calculate the probability of the complementary event, which is the probability that the coin will turn up heads on all three tosses, and then subtract it from 1.

The probability of getting heads on a single toss is 1/2. Since each toss is independent, the probability of getting heads on all three tosses is (1/2) * (1/2) * (1/2) = 1/8.

The probability of getting at least one tails is 1 - (probability of getting all heads) = 1 - 1/8 = 7/8.

Among the given answer choices, the probability 7/8 is represented by option (D).