If rate problems bring to mind moving trains, then there is no more iconic type of probability question than the coin toss. And there is good reason for this—coin tosses represent a fair portion of probability questions on the GRE. Even if a question doesn’t invoke the coin toss, the way we approach a coin toss problem can carry over to other types of probability questions.

## An Easy GRE Probability Question

*A fair-sided coin (which means no casino hanky-panky with the coin not coming up heads or tails 50% of the time) is tossed three times. What is the probability that I do not get two heads in a row? *

The way to approach this problem is by drawing out the problem. What we want to find is the number of conditions that satisfy the above requirement.

We can have one head in three different positions: HTT, TTH or THT.

We can also have two heads, but they cannot be consecutive, leaving us with HTH.

Were I to get all heads then that would break the restriction of not having at least two consecutive heads.

Finally, I can toss all tails: TTT.

So, in total I have five outcomes that satisfy the conditions in the question. We will call these the desired outcomes.

Once we’ve found this number, then what do we do?

With all probability questions, we always want to put the number of desired outcomes in the numerator. In the denominator, we want to put the total number of outcomes. The result of this fraction will give you the probability of an event happening.

### PROBABILITY = DESIRED OUTCOMES/POSSIBLE OUTCOMES

To find the total outcomes, we will have to calculate how many different outcomes can result from three tosses of a coin. Without writing them all out, we can calculate this number by multiplying the total different outcomes for each toss. Because we can only get heads or tails, the number of different outcomes for each toss is 2.

Therefore the total number of different outcomes for three tosses is 2 x 2 x 2 = 8. Returning to the equation above, we get 5/8.

Not too bad? Well, that was a warm-up. The probability that you’ll run into more difficult coin toss questions is very high. So, study this example and get comfortable with the process you need to follow to come to a correct probability every time.

any body help me for solving following problem

Seven coins are tossed. Given that there are

at least two heads. What is the (conditional)

probability of:

i. (exactly) 3 heads?

ii. At most 4 heads?

iii. No tails

Solve the above example by using sample

space and then by using the conditional probability.

Hi Chris, though this is a very old question, have a query regarding the paraphrasing of the question, specifically from here ” what is the probability that he/she is a girl who does not have blue eyes” …. now the moment we ask he/she we invariably are counting the boys in the count…which gives the answer as 29/32 … probably methinks that…if the question asks “what is the prob that the randomly chosen child is a girl who does not have blue eyes?” … this phrasing will make more sense…is it not ?

You’re absolutely right, Anup. Your wording works better than the somewhat awkward wording in R Kay’s original comment. Still, the actual problem R Kay posited is a pretty good one. 🙂

Hi Chris,

Please help me with this question. I am able to solve it by plugging in values but how do I solve it without that?

A weighted coin has a probability p of showing heads. If successive flips are independent, and the probability of getting at least one head in two flips is greater than 0.5, then what could p be?

Indicate all possible values.

[a] 0.1 [b] 0.2 [c] 0.3 [d] 0.4 [e] 0.6 [f] 0.7

Probability of Heads = p

Probability of Tails = (1-p)

Therefore the four possible outcomes have the following probabilities

Head, Head = p.p

Head, Tail = p(1-p)

Tail, Head = (1-p)

Tail, Tail = (1-p)^2

If the probability of getting atleast one head is greater than 0.5, then the probability of getting all tails is less than 0.5

Therefore we can come up with the following inequality.

(1-p)^2 < 0.5

We know that 0.5 is a little bigger than 0.7^2 which is 0.49, therefore

(1-p) 0.29

Therefore p can be 0.3, 0.4, 0.6, 07

c,d,e,f

Thanks Rasitha !! different approach than Mike !! It really help me understand it.

Another way to solve the question might be,

At least one heads in two flips = 1 – (Tails in both flips) = 1 – (1-p)^2

From the question

1- (1-p)^2 > 0.5

or, 0.5 > (1-p)^2

or, 0.71 > 1-p [taking positive root as probability can’t be negative]

or, p > 0.29

Hi ,

I have one doubt on below question,

A classroom has 12 girls and 20 boys. One quarter of the girls in the class have blue eyes. If a child is selected at

random from the class, what is the probability that he/she is a girl who does not have blue eyes?

can we solve it like, first we find probability of picking a girl having blue eye and then sustituting ans from 1… if yes then plz provide ne the solution???

k.Ray,

If there are 12 girls in the classroom and 1/4 have blue eyes, a total of 3 girls have blue eyes. That means that 9 or 12 girls do NOT have blue eyes.

So if a child is selected at random from class, the probability that child is a girl who does not have blue eyes equals: (Total # of girls without blue eyes)/(Total # of students) = 9/32.

Hope that helps!

hELLO, WHY CANT WE USE 1- (PROBABLITY OF GIRLS WHO HAVE BLUE EYES)…

Because if you use that, you’ll incorrectly get 29/32 instead of 9/32. the issue there is that 1 – (probability of girls who have blue eyes) doesn’t remove the probability of a randomly selected child being a boy.

Now, you COULD do 1 – (p of blue eyed girls) – (p of boys), but that may be a little more complicated than the method outlined above by Chris.

How did you know to multiply the outcomes of each coin toss instead of adding them?

Hi Clay,

The concept of multiplying relates to the fundamental counting principle. My colleague, Mike, in another post does a wonderful job of explaining it:

https://magoosh.com/gmat/2012/gmat-quant-how-to-count/

For the GRE (and the GMAT), you will definitely need to know how this concept.

Hope that post helps!

How can THT be counted when you said NO heads in two consecutive tosses?

Hi Ray,

By two consecutive heads, I meant the ‘head’ tosses cannot come back to back, as in HHT or THH. THT, however, only has one head. HTH would also be valid because the heads on the first toss is separated by the last heads by a tails on the second flip.

Hope that makes sense!

What about the case with all Tails,i.e. TTT ???

That should also be counted, making the probability 5/8?

Yep, you’re right :). Thanks for catching that!