soumya_joy wrote:A fair coin is tossed till a head appears for the second time. The probability that the
number of required tosses is 5, is
A. 1/16 B. 5/32 C. 1/8 D. 5/16

The 5-toss requirement is met when get exactly 1 head during the first 4 tosses, and then a head on the 5th toss.
There are 4 outcomes that meet this condition: HTTTH, THTTH, TTHTH, and TTTHH

So, we can write:
P(5 tosses required) = P(HTTTH OR THTTH OR TTHTH OR TTTHH)
= P(HTTTH) + P(THTTH) + P(TTHTH) + P(TTTHH)

As we'll soon see, the four separate probabilities above are all equal. To see why, let's examine one of them.
P(HTTTH) = P(Heads 1st AND Tails 2nd AND Tails 3rd AND Tails 4th AND Heads 5th)
= P(Heads 1st) x P(Tails 2nd) x P(Tails 3rd) x P(Tails 4th) x P(Heads 5th)
= 1/2 x 1/2 x 1/2 x 1/2 x 1/2
= 1/32

Using a similar technique, we can see that the other three probabilities also equal 1/2.

So, P(5 tosses required) = P(HTTTH OR THTTH OR TTHTH OR TTTHH)
= P(HTTTH) + P(THTTH) + P(TTHTH) + P(TTTHH)
= 1/32 + 1/32 + 1/32 + 1/32
= 4/32
= 1/8
= C

Cheers,
Brent

soumya_joy wrote:A fair coin is tossed till a head appears for the second time. The probability that the
number of required tosses is 5, is
A. 1/16 B. 5/32 C. 1/8 D. 5/16

We can also solve the question using counting techniques.

P(5 tosses) = [# of outcomes with exactly 1 head during the first 4 tosses, and then a head on the 5th toss]/[# of outcomes when coin is tossed 5 times]

As always, begin with the denominator.

# of outcomes when coin is tossed 5 times
Take the task of "building" outcomes and break it into stages.
Stage 1: Select outcome for 1st toss.
There are two options (H or T), so we can accomplish this stage in 2 ways.
Stage 2: Select outcome for 2nd toss.
We can accomplish this stage in 2 ways.
Stage 3: Select outcome for 3rd toss.
We can accomplish this stage in 2 ways.
Stage 4: Select outcome for 4th toss.
We can accomplish this stage in 2 ways.
Stage 5: Select outcome for 5th toss.
We can accomplish this stage in 2 ways.
By the Fundamental Counting Principle (FCP) we can complete all 5 stages (and thus "build" a 5-toss outcome) in (2)(2)(2)(2)(2) ways (= 32 ways)

# of outcomes with exactly 1 head during the first 4 tosses, and then a head on the 5th toss
Here, the 5th toss is fixed, since it must be Heads.
So, in how many ways can we select another toss to be Heads?
Well, we can have Heads on the 1st toss, the 2nd toss, the 3rd toss or the 4th toss.
So, there are 4 outcomes that meet our conditions (HTTTH, THTTH, TTHTH, and TTTHH)


So, P(5 tosses) = 4/32
= 1/8 = C

Cheers,
Brent

Aside: For more information about the FCP, we have a free video on the subject: https://www.gmatprepnow.com/module/gmat-counting?id=775