Probability

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knight247: Legendary Member; Posts: 504; Joined: Tue Apr 19, 2011 1:40 pmFollowed by: 11 members; Thanked: 114 times

by knight247 » Tue Sep 20, 2011 10:20 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)Can't be determined

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Source: Beat The GMAT — Problem Solving | Tue Sep 20, 2011 10:20 am

pemdas: Legendary Member; Posts: 1084; Joined: Fri Apr 15, 2011 2:33 pmFollowed by: 21 members; Thanked: 158 times

by pemdas » Tue Sep 20, 2011 10:41 am

revised, but not sure

if we consider 10 tosses with two clued (consecutive) attempts resulting in heads, then we have (10-1) consecutive attempts with heads out of 2^10 total possibilities <> 2*2*2 ... each time two options can be effective throughout ten attempts

two consecutive and eight other attempts are randomly chosen out of two options (head-tail), hence (2^8) for each of 9 attempts

to find P(0 consecutive attempts heads)= 1 - (2^8)/2^10 = 1 -1/4 = 3/4

d

knight247 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)Can't be determined

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gmatboost: Master | Next Rank: 500 Posts; Posts: 312; Joined: Tue Aug 02, 2011 3:16 pmFollowed by: 33 members; Location: New York City; Thanked: 130 times; GMAT Score:780

by gmatboost » Tue Sep 20, 2011 8:59 pm

This is extremely unrealistic and is not worth worrying about.

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phoenix111: Senior | Next Rank: 100 Posts; Posts: 47; Joined: Thu Jun 16, 2011 8:55 am; Location: San Francisco, CA; Thanked: 4 times

by phoenix111 » Tue Sep 20, 2011 10:17 pm

knight247 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)Can't be determined

Firstly, number of times 'heads' come should not be more than 5.( Otherwise, we will definetely have consecutive heads)

Number of ways when NO 'heads' come : 11c0 = 1
Number of ways when 1 'heads' come : 10c1 = 10
Number of ways when 2 'heads' come : 9c2 = 36
Number of ways when 3 'heads' come : 8c3 = 56
Number of ways when 4 'heads' come : 7c4 = 35
Number of ways when 5 'heads' come : 6c5 = 6

Total favourable outcome = 144
All possible = 2^10

Answer : 144/2^10 = 9/2^6 ( Option D. None of these )

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mankey: Legendary Member; Posts: 627; Joined: Thu Jun 23, 2011 9:12 amFollowed by: 1 member; Thanked: 4 times

by mankey » Wed Sep 21, 2011 4:30 am

Please explain this one.

Thanks
Mankey

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