If a coin is tossed 7 times then the total number of possible outcomes = 2*2*2*2*2*2*2 = 2^7
If we need x Heads out of these 7 tosses, we need to arrange these x identical Heads and (7 - x) identical Tails on the 7 available tosses. The total number of ways of that is 7!/(x!)(7 - x)! = 7Cx

Required Probability = P(0H,7T) + P(2H,5T) + P(4H,3T) + P(6H,1T)
where
P(0H,7T) = (7C0)/(2^7)
P(2H,5T) = (7C2)/(2^7)
P(4H,3T) = (7C4)/(2^7)
P(6H,1T) = (7C6)/(2^7)

Required probability
= (7C0 + 7C2 + 7C4 + 7C6)/(2^7)
= (1 + 21 + 35 + 7)/128
= 64/128
= 1/2

[spoiler](A)[/spoiler] is the answer

khandelwal.ab wrote:Hi I am not sure about this probability problem, can any expert help me here...

Q. A coin is tossed seven times, what is the probability that head appears even number of times?

A)1/2
b)1/4
c)1
d)1/3
E)None of the above

OA- A

Hi!

We can definitely solve using the coin flip formula and adding up the probability of each scenario that we want. However, we can solve MUCH quicker using logic.

We want all the cases with an even number of heads. So, we want:

0H
2H
4H
6h

We DON'T want all the cases with an odd number of heads. So, we don't want:

7H
5H
3H
1H

Now, if we're really on the ball, we see that each result that we do want has a symmetrical match with what we don't want. That is:

Prob(0H) = Prob(7H)
Prob(2H) = Prob(5H)
Prob(4H) = Prob(3H)
Prob(6H) = Prob(1H)

(How do we know that these are symmetrical? Well, there are a couple of ways to come to that realization:

1) notice that P(7H) = P(0T) and of course P(0T) = P(0H); we can do this for each case (P(2H) = P(5T) = P(5H), ...); or

2) understand that combinations are symmetrical and that combinations underlie coin flip questions. By symmetrical, we mean that nCk = nC(n-k); for example, 5C1=5C4; 12C4=12C8; 7C0=7C7...)

Since the scenarios we want are exactly equal to the scenarios we don't want, there must be a 50% chance of getting what we want... choose A!

khandelwal.ab wrote:Hi I am not sure about this probability problem, can any expert help me here...

Q. A coin is tossed seven times, what is the probability that head appears even number of times?

A)1/2
b)1/4
c)1
d)1/3
E)None of the above

OA- A

I received a PM asking me to comment.

On any given toss, P(heads) = 1/2.
When P=1/2, the probabilities exhibit SYMMETRY.

P(0 heads) = P(7 heads) [Since getting 7 heads is the same as getting 0 tails]
P(1 head) = P(6 heads) [Since getting 6 heads is the same as getting 1 tail]
P(2 heads = P(5 heads) [Since getting 5 heads is the same as getting 2 tails]
P(3 heads) = P(4 heads) [Since getting 4 heads is the same as getting 3 tails]

If we add together the equations -- placing the even outcome in each pair on the lefthand side and the odd outcome in each pair on the righthand side -- we get:
P(0 heads) + P(2 heads) + P(4 heads) + P(6 heads) = P(1 head) + P(3 heads) + P(5 heads) + P(7 heads).

Thus, P(even number of heads) = 1/2.

The correct answer is A.

Stuart/Mitch,

Thanks a lot for your responses. However I am still confused and unsure about the logic behind calculating the probability in each of the scenarios.

I am getting stuck at the fundamental step of calculating probability for each of the scenarios..

As per my current understanding, whatever scenario we may take the probability will be equal to (1/2)^7

For example consider 2 head and 4 tails..

then for 2 heads the probability is (1/2)^2, and for us to get the other 4 tails the probability will be (1/2)^5

Similarly if we consider any other scenario the probability will be (1/2)^7

I know that my approach and understanding is definitely incorrect.

It will be great if you guys could guide me!

Thanks again!

Hi,

I couldn't help chipping in here because this is such a common doubt.

For the case of 2 Heads and 5 Tails out of the 7 tosses, the outcomes favourable to us are:

H,H,T,T,T,T,T
T,T,T,T,T,H,H
H,T,H,T,T,T,T
T,H,H,T,T,T,T
..
and so on.

How many such cases will be there?
The number of ways will be same as the number of ways of arranging the letters (H, H, T, T, T, T, T), which is 7!/(5!*2!) or 7C2 or 7C5

Therefore, Number of favourable outcomes = 7C2
Total number of outcomes? 2*2*2*2*2*2*2 = 2^7

Method 1: By the definition of Probabilty,
Probability = (Number of favourable outcomes)/(Total number of outcomes)
So,
Required Probabilty = (7!/(5!*2!)/(2^7) or 7C2/(2^7)

General Case:
If there are r Heads and (n - r) Tails,
the Required Probabilty = (nCr)/(2^n)

Method 2: By adding up individual probability,
(and this is what you were trying to use)

For 2 Heads and 5 Tails,
Required Probability
= P(H,H,T,T,T,T,T) + P(H,T,H,T,T,T,T) + P(H,T,T,H,T,T,T) + .... P(T,T,T,T,T,H,H)
= (1/2)^7 + (1/2)^7 + (1/2)^7 + .... (1/2)^7
(How many terms do we have? As we saw above, it's 7C2)
= (7C2)*(1/2)^7

What you missed out is that the 7 tosses are all different and the 2 Heads and the 5 Tails can be arranged on them

If you understand this then Congratulations on getting an important concept clarified.

khandelwal.ab wrote:Stuart/Mitch,

Thanks a lot for your responses. However I am still confused and unsure about the logic behind calculating the probability in each of the scenarios.

I am getting stuck at the fundamental step of calculating probability for each of the scenarios..

As per my current understanding, whatever scenario we may take the probability will be equal to (1/2)^7

For example consider 2 head and 4 tails..

then for 2 heads the probability is (1/2)^2, and for us to get the other 4 tails the probability will be (1/2)^5

Similarly if we consider any other scenario the probability will be (1/2)^7

I know that my approach and understanding is definitely incorrect.

It will be great if you guys could guide me!

Thanks again!

Check my posts here:

https://www.beatthegmat.com/select-exact ... 88786.html

https://www.beatthegmat.com/probability- ... 91490.html

https://www.beatthegmat.com/rain-prob-t88396.html

khandelwal.ab wrote:Stuart/Mitch,

Thanks a lot for your responses. However I am still confused and unsure about the logic behind calculating the probability in each of the scenarios.

I am getting stuck at the fundamental step of calculating probability for each of the scenarios..

As per my current understanding, whatever scenario we may take the probability will be equal to (1/2)^7

For example consider 2 head and 4 tails..

then for 2 heads the probability is (1/2)^2, and for us to get the other 4 tails the probability will be (1/2)^5

Similarly if we consider any other scenario the probability will be (1/2)^7

I know that my approach and understanding is definitely incorrect.

It will be great if you guys could guide me!

Thanks again!

Hi!

Part of you're analysis is 100% correct: for any given string of 7 coin flips, there's a (1/2)^7 chance of it happening.

For example, there's a (1/2)^7 probability of getting:

HHHHHHH; or
HHHHHHT; or
HHHHHTT; and so on...

However, here's the problem: to satisfy what the question is asking, there's more than one acceptable string. Let's look at a simpler question to illustrate.

A fair coin is flipped twice. What's the probability of getting exactly 1 head and 1 tail?

Using your approach, the answer would be (1/2)^2 = 1/4.

However, there are two different ways to get 1 head and 1 tail:

HT
and
TH.

Since each of those strings has a 1/4 chance of happening, the answer to the question is:

1/4 + 1/4 = 1/2

So, here's the general rule for coin flip questions: to find the probability of getting k results on n flips, you divide the number of different strings that give you k results by the total number of possible strings.

Fortunately, there's a formula that makes life easy!

The probability of getting k results in n flips is:

nCk/2^n

(as a reminder, nCk is the combinations formula: n!/k!(n-k)!).

For more on coin flip questions, check out:

https://www.beatthegmat.com/coin-flip-qu ... html#75414