alex.gellatly wrote:If a coin has an equal probability of landing heads up or tails up each time it is flipped, what is the probability that the coin will land heads up exactly twice in 3 consecutive flips?

(A) 0.125
(B) 0.25
(C) 0.375
(D) 0.50
(E) 0.666

Thanks

3* (1/2)^3 = 3/8 = 0.375
Hence C

killer1387 wrote:

alex.gellatly wrote:If a coin has an equal probability of landing heads up or tails up each time it is flipped, what is the probability that the coin will land heads up exactly twice in 3 consecutive flips?

(A) 0.125
(B) 0.25
(C) 0.375
(D) 0.50
(E) 0.666

Thanks

3* (1/2)^3 = 3/8 = 0.375
Hence C

Care to explain?

Head comes twice, and Tail comes one.
But the order of the Heads and the Tail is not mentioned.
You can have a head,head,tail OR head,tail,head OR tail,head,head.
Reqd. Probability = P(H,H,T) or P(H,T,H) or P(T,H,H)
= 0.5*0.5*0.5 + 0.5*0.5*0.5 + 0.5*0.5*0.5
= 3/8

But, the above way is lengthy when there are too many ARRANGEMENTS of the Heads and the Tails. The below is a better way:

Instead of writing all the cases for two heads and one tail, we can arrange 2Hs and 1T in 3!/2! ways and multiply (1/2)*(1/2)*(1/2) by the number of arrangements.

Therefore,
required probability = (1/2)*(1/2)*(1/2)*(3!/2!)
= 3/8

Alternatively:
Probability = Favourable Events/Total no. of events
Total number of possibilities for 3 tosses = 2*2*2
Favourable events (are all the arrangements of 2Hs and 1T) = 3!/2!
therefore,
required probability = (3!/2!)/(2^3)
= 3/8

Practice Problem:
If a coin is tossed 5 times, what is the probability that:
(a) you get 3 Heads?
(b) you get atleast 4 Heads?

aneesh.kg wrote:Head comes twice, and Tail comes one.
But the order of the Heads and the Tail is not mentioned.
You can have a head,head,tail OR head,tail,head OR tail,head,head.
Reqd. Probability = P(H,H,T) or P(H,T,H) or P(T,H,H)
= 0.5*0.5*0.5 + 0.5*0.5*0.5 + 0.5*0.5*0.5
= 3/8

But, the above way is lengthy when there are too many ARRANGEMENTS of the Heads and the Tails. The below is a better way:

Instead of writing all the cases for two heads and one tail, we can arrange 2Hs and 1T in 3!/2! ways and multiply (1/2)*(1/2)*(1/2) by the number of arrangements.

Therefore,
required probability = (1/2)*(1/2)*(1/2)*(3!/2!)
= 3/8

Alternatively:
Probability = Favourable Events/Total no. of events
Total number of possibilities for 3 tosses = 2*2*2
Favourable events (are all the arrangements of 2Hs and 1T) = 3!/2!
therefore,
required probability = (3!/2!)/(2^3)
= 3/8

Practice Problem:
If a coin is tossed 5 times, what is the probability that:
(a) you get 3 Heads?
(b) you get atleast 4 Heads?

Are the answers to your practice problems as follows:
A. 5 / 32
B. 4 / 32
Thanks

This is a permutation question as the ordering is important. All the three ways mentioned are important in this question.

The practice problem's solution:

A] 9/32 as the order is not confirm, we shall have 3 possibilities as follows:

1. {H, H, H, T, T}

2. {T, T, H, H, H}

3. {T, H, H, H, T}

4. {T, H, T, H, H}

5. {T, H, H, T, H}

6. {H, T, H, H, T}

7. {H, T, T, H, H}

8. {H, H, T, T, H}

9. {H, T, H, T, H}


B] The answer is 3/16 . The probability of getting 4 heads at least is as follows:


1. {H, H, H, H, T}

2. {H, H, H, T, H}

3. {H, H, T, H, H}

4. {H, T, H, H, H}

5. {T, H, H, H, H}

6. {H, H, H, H, H}

In all 6 ways and all of them having a probability of 1/2 hence 6/32 = 3/16.


I really hope I am correct. As per the formula this is what it should be...

Alex: Both your answers are wrong. Read the explanation given before and try the problem again before reading the solution below.
Amit: You're are counting all the cases, which is tedious. That's why you made a mistake in the first part of the problem. Read the explanation given before and read the solution below.

P (3 Heads) = (1/2)*(1/2)*(1/2)*(1/2)*(1/2)*(5!/3!2!)
= 10/32 = 5/16

P (atleast 4 Heads) = P(4 Heads) + P(5 Heads)
= (1/32)^5 * (5!/4!) + (1/32)^5 * 1
= (5 + 1)/32
= 6/32
= 3/16