probability

[sud21]

A fair coin is tossed 5 times. What is the probability of getting at least three heads on consecutive tosses?

[Mike@Magoosh]

Hi, there. I'm happy to contribute to this.

This is a relatively challenging probability question.

First of all, toss a fair coin five times. Each toss has two possible outcomes, so the total number of five-toss outcomes is 2^5 = 32.

How many of those involve getting at least three heads on consecutive tosses?

Case 1: 3 H and 2 T

Here, the cases that satisfy the condition are:
HHHTT, THHHT, and TTHHH ----> 3 cases

Case 2: 4 H and 1 T

Here, the cases that satisfy the condition are:
HHHHT, HHHTH, HTHHH, HHHHT ----> 4 cases

Case 3: 5 H and 0 T

Here, there's only one case -- HHHHH -- and it satisfies the condition.

3 + 4 + 1 = 8 cases total.

P(at least three heads on consecutive tosses) = 8/32 = 1/4

Does that make sense? Please let me know if you have any questions.

Mike

[Anurag@Gurome]

A fair coin is tossed 5 times. What is the probability of getting at least three heads on consecutive tosses?

At least 3 heads implies it can 3 or 4 or 5 heads.

Case 1: Possibility of getting 3 heads on consecutive tosses
HHHTT, TTHHH, THHHT, HTHTH, HTHHH, HHHTH
So, the probability of getting 3 heads on consecutive tosses = 5 * (1/2)^5 = 5/32

Case 2: Possibility of getting 4 heads on consecutive tosses
HHHHT, THHHH
So, the probability of getting 4 heads on consecutive tosses = 2 * (1/2)^5 = 2/32

Case 3: Possibility of getting 5 heads on consecutive tosses
HHHHH
So, the probability of getting 5 heads on consecutive tosses = 1 * (1/2)^5 = 1/32

Therefore, required probability = 5/32 + 2/32 + 1/32 = 8/32 = [spoiler]1/4[/spoiler]

[spoiler][/spoiler]

[gmatdriller]

How does BINOMIAL EXPANSION work here?

(H + T)^5 = H^5 + 5H^4*T + 10H^3*T^2 + 10H^2*T^3 + 5H*T^4 + T^5

I feel we could collect just the terms that correspond to the number
of outcomes requested.

For example: An outcome of 4 Heads, 1 Tail => 5H^4*T
This is equal 5(1/2^5) = 5/32

But for an outcome of 3Heads, 2Tails = 10H^3*T^2 = 10/32 ..however, this is wrong
Correct answer = 5(favorable outome) / 32 (Total possible outome)

CAN SOMEONE PLEASE CORRECT WHERE I HAVE GONE WRONG

[Mike@Magoosh]

How does BINOMIAL EXPANSION work here?

(H + T)^5 = H^5 + 5H^4*T + 10H^3*T^2 + 10H^2*T^3 + 5H*T^4 + T^5

I feel we could collect just the terms that correspond to the number
of outcomes requested.

For example: An outcome of 4 Heads, 1 Tail => 5H^4*T
This is equal 5(1/2^5) = 5/32

But for an outcome of 3Heads, 2Tails = 10H^3*T^2 = 10/32 ..however, this is wrong
Correct answer = 5(favorable outome) / 32 (Total possible outome)

CAN SOMEONE PLEASE CORRECT WHERE I HAVE GONE WRONG

Dear gmatdriller,
I'm happy to respond.

I believe you overlooked the crucial word "consecutive". If the question were just about 3+ heads in five tosses, then order would not matter at all, and the binomial expansion would be perfectly appropriate.

Instead, because we are concerned with 3+ consecutive heads, order means everything, and the binomial expansion is useless.

Does this make sense?
Mike

[gmatdriller]

Hello Mike,
Here the Heads have to follow each other without interruption
in any Tails, right?

In that case, the example given by Sund21(shown below) would be wrong:
"Case 1: Possibility of getting 3 heads on consecutive tosses
HHHTT, TTHHH, THHHT, HTHTH, HTHHH, HHHTH"

I can see that if order matters, we would also have fewer no of cases.
In Binomial(order is not relevant)we have several pattern of 10.
In consecutive, there are only 5 ceases.

Please comment if my explanations are correct,

Thanks

Please

[Mike@Magoosh]

Hello Mike,
Here the Heads have to follow each other without interruption
in any Tails, right?

In that case, the example given by Sund21(shown below) would be wrong:
"Case 1: Possibility of getting 3 heads on consecutive tosses
HHHTT, TTHHH, THHHT, HTHTH, HTHHH, HHHTH"

I can see that if order matters, we would also have fewer no of cases.
In Binomial(order is not relevant)we have several pattern of 10.
In consecutive, there are only 5 ceases.

Please comment if my explanations are correct,

Thanks

Please

Dear gmatdriller,
I think you misunderstand the exact meaning of the word "consecutive." The condition stated in the problem is "What is the probability of getting at least three heads on consecutive tosses?" In other words, three consecutive tosses, three tosses in a row, all have to be H. The string of five tosses must contain, somewhere, a streak of three H's in a row. As long as there are three H's a row, it doesn't matter if the others are head or tails and whether they are adjacent to the streak of three. For example, here are some scenarios that meet the condition:
HHHTT = YES
HHHTH = YES
THHHT = YES
TTHHH = YES
HTHHH = YES
Each one of those has a streak of three consecutive H's, so those correctly satisfy the condition. Of course, the condition allows for more than three consecutive heads, because of those important words "at least", so longer streaks are allowed:
HHHHT = YES
THHHH = YES
HHHHH = YES
Those eight cover all the possibilities in which a series of five coin tosses would have three or more consecutive H's. If the H's are split us, such that there are no streaks of three, then those don't meet the condition. For example
HTHTH = NO
HHTHH = NO
Both of those have 3 or more H's, but there is not a streak of three in a row, so they do not meet the condition.

Does all this make sense?

Mike

[gmatdriller]

Thanks Mike for taking the time to explain further.

You seem, however, to explain just what I stated in my post.:

"HTHTH" is NOT a string of at least 3 consecutive Heads

HHHTT has 3 consecutive Heads with no interruption until after
the 3rd H.

In sum, Binomial expansion includes both the consecutives and
non-consecutives. And so, the Binomial option (10H3T^2) has more terms.

Thanks so much Mike.