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

Awesome explanation Bunuel !

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

Bunuel - U rock.
+1 for you.

Cheers.!

This is like another probability question asked in the forum, bunnel explained that very well.

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.

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.

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.

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.

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

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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?

(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, ...

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

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