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tossed 10 times

by sanju09 » Fri Feb 20, 2009 3:15 am
A fair coin is tossed 10 times. What is the probability that two heads do not occur consecutively?

(A) 1/ 2^4

(B) 1/2^3

(C)1/2^5

(D) None of the above

(E) It cannot be determined by the information provided
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Re: tossed 10 times

by billzhao » Fri Feb 20, 2009 4:18 am
sanju09 wrote:A fair coin is tossed 10 times. What is the probability that two heads do not occur consecutively?

(A) 1/ 2^4

(B) 1/2^3

(C)1/2^5

(D) None of the above

(E) It cannot be determined by the information provided
The answer is 1/2+1/2^10.

The probability that two heads occur consecutively = (1/2)^2+(1/2)^3+(1/2)^4+...+(1/2)^10=1/2-1/2^10
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Re: tossed 10 times

by sanju09 » Fri Feb 20, 2009 5:16 am
billzhao wrote:
sanju09 wrote:A fair coin is tossed 10 times. What is the probability that two heads do not occur consecutively?

(A) 1/ 2^4

(B) 1/2^3

(C)1/2^5

(D) None of the above

(E) It cannot be determined by the information provided
The answer is 1/2+1/2^10.

The probability that two heads occur consecutively = (1/2)^2+(1/2)^3+(1/2)^4+...+(1/2)^10=1/2-1/2^10
:shock: This problem probably needs more effort than you have already put. I am leaving this question to public with only two clues.

1. billzhao didn't explain it correctly.

2. This is a 720+ treat.

Over to anybody now...
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Re: tossed 10 times

by masuarezdl » Fri Feb 20, 2009 8:57 am
sanju09 wrote:
billzhao wrote:
sanju09 wrote:A fair coin is tossed 10 times. What is the probability that two heads do not occur consecutively?

(A) 1/ 2^4

(B) 1/2^3

(C)1/2^5

(D) None of the above

(E) It cannot be determined by the information provided
The answer is 1/2+1/2^10.

The probability that two heads occur consecutively = (1/2)^2+(1/2)^3+(1/2)^4+...+(1/2)^10=1/2-1/2^10
Hmmm... I would go with answer D.

For the first flip, it can be any of two coins, thus 2/2. For the next ones, it has to be 1/2, since we have one particular side that we need in order for it to be different from the previous. Therefore, the equation:

(2/2)*(1/2)*(1/2)*(1/2)*(1/2)*(1/2)*(1/2)*(1/2)*(1/2)*(1/2)=(2/(2^10))

Any thoughts?
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by 4meonly » Fri Feb 20, 2009 10:45 am
I am also with D
A fair coin is tossed 10 times. What is the probability that two heads do not occur consecutively?
Allowed combinations = total combinations - restricted combinations
Prabability A not occur = 1 - Probability A occur

Total combinations = 2^10

Let 2 heads (HH) will be X

Restricted combinations
With 1 pair HH = XTTTTTTTT = 9!/8! = 9 combinations
With 2 pairs HH = XXTTTTTT = 8!/2!6! = 28 combinations
With 3 pairs HH = XXXTTTT = 7!/3!4! = 35 combinations
With 4 pairs HH = XXXXTT = 6!/4!2! = 15 combinations
With 5 pairs HH = XXXXX = 5!/5! = 1 combination
Total restricted combinations = 88

Allowed combinations = 2^10 - 88 = 1024 - 88 = 936


Probability = necessary combinations / total combinations
936/1024 = 117/128
Hence, D

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Heads or Tails

by kanha81 » Fri Feb 20, 2009 1:13 pm
I do not know if I have solved it correctly or not, but this is how I conclude that the answer is [D]

q: P(No 2 Heads occur together) = ?
P(H)=1/2
P(T)=1/2

Listing all winning scenarios
toss(1)=H, toss(2)=T, toss(3)=H...and so on: (1/2)^10
toss(1)=T, toss(2)=H, toss(3)=T...and so on: (1/2)^10
toss(1)=T, toss(2)=T, toss(3)=T...and so on: (1/2)^10
toss(1)=H, toss(2)=T, toss(3)=T...and so on: 2(1/2)^10
toss(1)=H, toss(2)=T, ..toss(10)=H : (1/2)^10

4*(1/2)^10 + 2*(1/2)^10 = 6*(1/2)^10 = 3*(1/2)^9

is it correct? what would be the most efficient way?

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by earth@work » Fri Feb 20, 2009 1:53 pm
D here too

HTHTHTHTHT ...1 way (5H)
THTHTHTHTH ....1 way (5H)
TTTTTTTTTH ......10 way 1H)
HTHTTTTTTT .......8ways (2H)-- Error! 7*8 +8*2 ways=72ways...TTHTHTTTTT etc were not taken into account... similarly error in rest below! I dnt think tis is the correct method to solve it : there should be a shorter method!!
THTHTTTTTT .......7ways(2H)
HTHTHTTTTT........6way(3H)
THTHTHTTTT.......5ways(3H)
HTHTHTHTTT.......4ways(4h)
THTHTHTHTT......3ways(4H)
TTTTTTTTTT .......1 WAY(0H)

Total fav. ways = 2+3+4+5+6+7+8+10+1=46
total ways = 2^10
Prob =46/(2^10) =23/2^9...... D
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by sanju09 » Sat Feb 21, 2009 2:05 am
Let's see if this clue helps for a better and shorter explanation to this problem:

The number of words you can make are just numbers from the Fibonacci sequence, since a_1 = 1 and a_2 = 2. If you did this problem for any number of coins, the numerators would all be Fibonacci numbers. So for 2, 3, 4, 5, 6, 7, 8, 9, and 10 coins, the answers will be:

3/4; 5/8; 8/16; 13/32; 21/64; 34/128; 55/256; 89/512; 144/1028...

:) OA should be D.
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by 4meonly » Sat Feb 21, 2009 2:32 pm
sanju09 wrote:Let's see if this clue helps for a better and shorter explanation to this problem:

The number of words you can make are just numbers from the Fibonacci sequence, since a_1 = 1 and a_2 = 2. If you did this problem for any number of coins, the numerators would all be Fibonacci numbers. So for 2, 3, 4, 5, 6, 7, 8, 9, and 10 coins, the answers will be:

3/4; 5/8; 8/16; 13/32; 21/64; 34/128; 55/256; 89/512; 144/1028...

:) OA should be D.
HMMMMM, to whom are you speaking? WTF? :? :shock: :lol:

Could you please do it in a simnlier way? :D
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