A fair coin is tossed 5 times. What is the probability of getting at

Question

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

A. 2/16
B. 1/4
C. 7/24
D. 5/16
E. 15/32

Bunuel · Accepted Answer

At least 3 heads means 3, 4, or 5 heads.

3  consecutive  heads 
5 cases:
HHHTT
THHHT
TTHHH
HTHHH
HHHTH

P=5*(\frac{1}{2})^5=\frac{5}{32} .

4  consecutive  heads 
2 cases:
HHHHT
THHHH

P=2*(\frac{1}{2})^5=\frac{2}{32} .

5  consecutive  heads 
1 case:
HHHHH

P=(\frac{1}{2})^5=\frac{1}{32} .

P=\frac{5}{32}+\frac{2}{32}+\frac{1}{32}=\frac{8}{32}=\frac{1}{4} .

Answer: B.

SVaidyaraman · Answer

In this type of problems where repetition of values is allowed, the total number of possibilities is given by the formula  n^r  where n is 2 and has the values Heads and Tails. r is 5 and is equal to the number of tosses.

1. Total number of possibilities =  2^5 = 32 
2. Instances of favorable outcomes:

(i) HHH** - Let us elaborate all the possibilities:

1. HHH HH
2. HHH HT
3. HHH TH
4. HHH TT

We have exhausted the possibilities.

(ii) We also have the following **HHH and the possibilities are:

5. HH HHH
6.HT HHH
7,TH HHH
8.TT HHH

(iii) and the following: *HHH*

9. H HHH H
10, H HHH T
11. T HHH H
12. T HHH T

Out of the total 12 , only 8 are unique.

3. The probability is  8/32 = 1/4.

qweert · Answer

Brunnel explained it in detail. I would just count the number of possibilities and divide it by 2^n

HHHHH
HHHTT
HHHHT
THHHT
TTHHH
HHHTH
THHHH
HTHHH

8/(2)^5 ==> 8/32 ==> 1/4

gmatbull · Answer

I prefer using Binomial Expansion:

(H + T)^5: H^5 + 5H^4(T) +  10H^3*T^2  + 10H^2*T^3 + 5H*T^4 + T^5
The outcome for 3Heads 2 Tails = 10H^3*T^2 => 10*(1/32) = 5/16

The correct value should be 5/32

Can someone correct what am missing out here?

Thanks.

MartyTargetTestPrep · Answer

Yes. It's not just three heads two tails. Its three heads in a row, AND, it could also be four heads in a row or five heads in a row.

AryamaDuttaSaikia · Answer

In this question, i find "Counting" the cases is the best way to solve.

However, to answer the last question posted : In the combination (as is used by you for 3 consecutive heads) case, we are taking 3 heads in five tosses and not 3 consecutive heads in as many tosses.

SVaidyaraman · Answer

A general solution for constraints problems with line arrangement :

1. Represent the constraints at the left most. The first constraint is 3 consecutive heads
HHH_ _
2. Fill up the right most blanks 
It can be TT and so HHHTT
3. Shift together what is constrained, one position at a time to the right and count the number of valid cases.
Here they are HHHTT, THHHT, TTHHH. So there are 3 cases
4. See if what is not constrained can be reordered. 
In this case it is TT and cannot be reordered.
If it can be reordered, repeat step 3.
5. See if what is not constrained can change
If can change, make the change and go to step 2 else stop.
Here TT can change to TH or HT. Let us take TH.

Finally, we get the valid cases HHHTH and HTHHH. So there are 2 cases

The total number of cases for 3 consecutive Heads is 5.

Similarly we can do for the constraints 4 and 5 consecutive heads. 
Note: When there are many constraints such as 3,4 or 5 consecutive heads start with the least constrained i.,e 3 here.

push12345 · Answer

More than 3 consecutive heads

3 heads 2 tails
5!/(3!2!)=10 for consecutive cases are half so 5 cases

4 head 1 tail
5!/(4!1!)=5 ,for consecutive cases r half so 2 cases

5 head 
1 case only

Total valid cases=8

Total cases=2*2*2*2*2 ( as each coin can take either H or T)

Result= 8/32=1/4

Posted from my mobile device

rainingrelics · Answer

Bunuel 
In this case we why can't we follow this method -
3 Heads, 2 Tails
(1/2)^35 * 5!/3!2! = 10/35

4 Heads, 1 Tail
(1/2)^35 * 5!/4! = 5/35

5 Heads
(1/2)^35 = 1/35

Total Probability = 16/35.

Why is this method wrong?

Bunuel · Answer

(1/2)^5 * 5!/(3!2!) gives all arrangements of HHHTT, not only those in which we have 3  consecutive  heads. For example, HHTTH, HTHTHH, ...

BrushMyQuant · Answer

Given that A fair coin is tossed 5 times and we need to find What is the probability of getting at least three heads on consecutive tosses

At least 3 heads means that we can get 3 or more heads in consecutive tosses.

=>  P(At least 3 consecutive heads) = P(3 consecutive heads) + P(4 consecutive heads) + P(5 consecutive heads)

P(3 consecutive heads)

Total number of cases =  2^5  = 32
Case in which we get 3 consecutive heads are HHHTT, HHHTH, THHHT, TTHHH, HTHHH => 5

=> P(3 consecutive heads) =  \frac{5}{32}

P(4 consecutive heads)

Case in which we get 4 consecutive heads are HHHHT, THHHH => 2

=> P(4 consecutive heads) =  \frac{2}{32}

P(5 consecutive heads)

Case in which we get 5 consecutive heads are HHHHH => 1

=> P(5 consecutive heads) =  \frac{1}{32}

=>  P(At least 3 consecutive heads) = P(3 consecutive heads) + P(4 consecutive heads) + P(5 consecutive heads)  =  \frac{5}{32}  +  \frac{2}{32}  +  \frac{1}{32}  =  \frac{8}{32}  =  \frac{1}{4}

So,  Answer will be B 
Hope it helps!

Watch the following video to learn How to Solve Probability with Coin Toss Problems

https://www.youtube.com/watch?v=LRURv38kZw8

hamzanoman1212 · Answer

Bunuel

Hi,

Can you please help me understand why my approach is wrong?

For HHHTT (3 heads 2 tails), I am thinking of HHH as one number i.e. H1,T,T. And there's three ways to arrange it = 3! = 6. However, the correct number when I write down the sets is 5, similar to what you've done.

Any help would be appreciated.

Thanks

pick51 · Answer

I also have the same question, why is this wrong?

Bunuel · Answer

First, if you consider {HHH} as one unit, the number of ways to arrange {HHH}, {T}, {T} will be 3!/2! = 3, not 3! = 6:

{HHH}{T}{T} 
{T}{HHH}{T} 
{T}{T}{HHH}

Next, the above three cases miss two additional cases where three consecutive heads, {HHH}, exist alongside a single separate head, {H}:
{HHH}{T}{H} 
 {H}{T}{HHH}

pick51 · Answer

Thank you for your reply, that makes sense! Is there an "elegant" way to count up the number of possibilities for this problem, or is the only way to just manually go through the possibilities?

Bunuel · Answer

If I knew a more "elegant" way, I’d share it. I’m not withholding better methods than I post—if I’m aware of them, I’ll share them.

adityagupta0400 · Answer

Let's consider this on a case-by-case basis-

1. We have only 3 consecutive heads-
 a. 3H and 2T-
 Since 3H are together, we have 3 positions to select from to place 3H
 _ _ _ -> We have to choose a position to place 3H -> 3C1 =  3

b. 3H, 1H and 1T-
 _ _ _ -> 3H and 1H can't be together (otherwise it would be 4 consecutive Hs) -> This means 1T will have to be in the middle
 Which means, we have 2 positions to select from (left and right) -> 2C1 =  2

2. We have 4 consecutive heads-
 4H and 1T
 Since 4H are together, we have 2 positions to fill- 
 _ _ -> 2C1 =  2

3. We have 5 consecutive heads- 
 Only 1 way to do this ->  1

Total ways to have 3 consecutive heads = 3 + 2+ 2 + 1 = 8.

Total number of possibilities = 2^5 = 32.

Probability of getting at least 3 consecutive heads = 8/32 =  1/4 .

A fair coin is tossed 5 times. What is the probability of getting at

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A fair coin is tossed 5 times. What is the probability of getting at

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