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by daretodream » Fri Feb 19, 2010 3:26 am
The probability is 1/2 that a coin will turn out to be tails on any given toss. A coin is tossed 5 times. What is the probability of getting at least three heads on consecutive tosses.
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by ajith » Fri Feb 19, 2010 3:54 am
daretodream wrote:The probability is 1/2 that a coin will turn out to be tails on any given toss. A coin is tossed 5 times. What is the probability of getting at least three heads on consecutive tosses.
If we consider the block of three consecutive heads as H

Total out comes = 32
Favorable outcomes = All Heads,4 consecutive Heads and tail, Three consecutive heads and two tails

=1+2+3 =6

Probability = 3/16
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by harsh.champ » Fri Feb 19, 2010 4:41 am
daretodream wrote:The probability is 1/2 that a coin will turn out to be tails on any given toss. A coin is tossed 5 times. What is the probability of getting at least three heads on consecutive tosses.
In such questions ,it is important to note that in any case the probability will be (1/2) only.[whether it is heads or tails]
So as a shortcut approach many a times you will find that the answer is (1/2)^x (where x is the no. of the tosses.)

Now, over here we are talking about consecutive tosses ,so we also have to consider the selections of the order of those tails.

Hence,Answer will be [(1/2)^x]*(No. of arrangements)

Case 1:-All Heads.Only 1 selection.
Case 2:-4 consecutive Heads and tail.(The posn. of tail can be selected in 2 ways)
Case 3:-Three consecutive heads and two tails(the 2 tails can be selected in 3 ways).

Hence,the [spoiler]answer will be [(1/2)^x]*(No. of arrangements) = [(1/2)^5]*(6)
=6/32 =3/16[/spoiler]
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by vijay_venky » Fri Feb 19, 2010 5:40 am
I think the answer is 1/4.

1. 3 heads and 2 tails

HHHTT
THHHT
TTHHH

2. 4 heads and 1 tail

HHHHT
HHHTH
HTHHH
THHHH

3. all 5 heads

HHHHH

So a total of 8 probabilities with a sample space of 32. And the probability is 1/4.
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by harsh.champ » Fri Feb 19, 2010 6:00 am
vijay_venky wrote:I think the answer is 1/4.

1. 3 heads and 2 tails

HHHTT
THHHT
TTHHH

2. 4 heads and 1 tail

HHHHT
HHHTH
HTHHH
THHHH

3. all 5 heads

HHHHH

So a total of 8 probabilities with a sample space of 32. And the probability is 1/4.
Hey venky,
Thanks we left out the following 2 cases:-
HHHTH
HTHHH

In the question it was written that 3 heads have to be consecutive,not 4.
So,the 4th head can be separate also.We missed 2 alternatives in the 2nd case- 4 heads and 1 tail.
This was a very good trap.
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by shashank.ism » Fri Feb 19, 2010 7:14 am
vijay_venky wrote:I think the answer is 1/4.

1. 3 heads and 2 tails

HHHTT
THHHT
TTHHH

2. 4 heads and 1 tail

HHHHT
HHHTH
HTHHH
THHHH

3. all 5 heads

HHHHH

So a total of 8 probabilities with a sample space of 32. And the probability is 1/4.
venky that was a good idea to approach by checking all the probabilities ...Otherwise we miss out 1 or 2 options as harsh and ajith...

very nice question daretodream .. I always happen to do mistakes in probability ..due to such tricks involved in it.
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by kstv » Fri Feb 19, 2010 9:34 am
Probability of three H in a Row is 1/2 X 3 or 1/8
This can happen in three ways
HHHTT, THHHT, TTHHH

so 1/8 X 3 = 3/8

Prob of 4 H is 2/16

Prob of 5 H is 1/32


Prob. of atleast 3 Heads in a Row = 3/8+2+16 + 1/32 = 17/32
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