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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 least three heads on consecutive tosses? A. 2/16 B. 1/4 C. 7/24 D. 5/16 E. 15/32
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Originally posted by nusmavrik on 20 Aug 2010, 05:39.
Last edited by Bunuel on 05 Feb 2019, 05:21, edited 1 time in total.
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Re: A fair coin is tossed 5 times. What is the probability of getting at
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20 Aug 2010, 05:59
nusmavrik wrote: 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 At least 3 heads means 3, 4, or 5 heads. 3 consecutive heads5 cases: HHHTT THHHT TTHHH HTHHH HHHTH \(P=5*(\frac{1}{2})^5=\frac{5}{32}\). 4 consecutive heads2 cases: HHHHT THHHH \(P=2*(\frac{1}{2})^5=\frac{2}{32}\). 5 consecutive heads1 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.
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Re: A fair coin is tossed 5 times. What is the probability of getting at
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21 Aug 2010, 00:04
Awesome explanation Bunuel !



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Re: A fair coin is tossed 5 times. What is the probability of getting at
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21 Aug 2010, 00:50
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



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Re: A fair coin is tossed 5 times. What is the probability of getting at
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21 Aug 2010, 13:54
All possible events = (1/2)^5= 1/32 all the favorable events HHHHH HHHTT THHHT TTHHH HHHHT THHHH that makes 6 favourable events Hence answer will 6/32 = 1/4 therefore B
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Re: A fair coin is tossed 5 times. What is the probability of getting at
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23 Aug 2011, 17:11
my question is how do you determine from this problem, that immediately, you are going to have to map out the possibilities?
there seems to be no nifty way to do it other than really just exhausting the possibilities?
thanks



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Re: A fair coin is tossed 5 times. What is the probability of getting at
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23 Aug 2011, 17:51
Bunuel  U rock. +1 for you. Cheers.!
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Re: A fair coin is tossed 5 times. What is the probability of getting at
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27 Dec 2011, 05:23
This is like another probability question asked in the forum, bunnel explained that very well.



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Re: A fair coin is tossed 5 times. What is the probability of getting at
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24 Mar 2012, 11:53
A fair coin is tossed 5 times. What is the probability of getting at least three heads on consecutive tosses? A. 3/16 B. 1/4 C. 7/24 D. 5/16 E. 15/32 Kaplan says the answer is 1/4. I think the answer is 3/16 because I have interpreted the qn as: What is the probability of getting atleast 3 consecutive heads > 3 consecutive heads or 4 consecutive heads or 5 consecutive heads ?



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Re: A fair coin is tossed 5 times. What is the probability of getting at
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24 Mar 2012, 14:29
shahlanuk wrote: A fair coin is tossed 5 times. What is the probability of getting at least three heads on consecutive tosses? A. 3/16 B. 1/4 C. 7/24 D. 5/16 E. 15/32 Kaplan says the answer is 1/4. I think the answer is 3/16 because I have interpreted the qn as: What is the probability of getting atleast 3 consecutive heads > 3 consecutive heads or 4 consecutive heads or 5 consecutive heads ? Merging similar topics. Your interpretation is correct: at least 3 consecutive heads means 3, 4, or 5 consecutive heads, but the answer is still 1/4. Please check the solution provided above and ask if anything remains unclear.
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Re: A fair coin is tossed 5 times. What is the probability of getting at
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13 Nov 2012, 21:31
bellcurve wrote: A fair coin is tossed 5 times, what is the probability of getting at least 3 heads on consecutive tosses?
 Ans will be provided later. Favorable outcomes: HHHHH  All heads HHHHT  4 heads THHHH HHHTT  3 heads THHHT TTHHH HTHHH HHHTH Total possiblities = 2^5 = 32 Favorable = 8 Probability = 8/32 = 1/4 Hope it helps! PS: Please provide atleast options if not OA.
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Re: A fair coin is tossed 5 times. What is the probability of getting at
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13 Nov 2012, 23:02
at least consecutive 3 heads max consecutive 5 heads
5 heads  HHHHH (1 way) 4 heads  THHHH, HHHHT (2 ways) 3 heads  HHHTH, HHHTT, THHHT, HTHHH, TTHHH (5 ways)
Total 8 ways
Total no ways irrespective of outcome= 2x2x2x2x2 = 32
probability of atleast 3 consecutive heads = 8/32 or .25



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Re: A fair coin is tossed 5 times. What is the probability of getting at
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15 Jul 2013, 07:03
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 \(8/32 = 1/4.\)
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Re: A fair coin is tossed 5 times. What is the probability of getting at
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16 Mar 2014, 09:30
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.
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Re: A fair coin is tossed 5 times. What is the probability of getting at
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11 Jan 2015, 06:41
gmatbull wrote: 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? 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.
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Re: A fair coin is tossed 5 times. What is the probability of getting at
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09 Mar 2015, 06:06
I missed two cases HHHTH and HTHHH . Great solution by Bunnel.
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Re: A fair coin is tossed 5 times. What is the probability of getting at
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23 May 2015, 00:57
Can someone suggest another way of solving this without writing down the possibilities? Maybe use combinations.An alternative could be useful in case you miss out on a case while jotting down the possibilities..thanks



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Re: A fair coin is tossed 5 times. What is the probability of getting at
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20 Sep 2015, 21:56
Bunuel wrote: nusmavrik wrote: 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 At least 3 heads means 3, 4, or 5 heads. 3 consecutive heads5 cases: HHHTT THHHT TTHHH HTHHH HHHTH \(P=5*(\frac{1}{2})^5=\frac{5}{32}\). 4 consecutive heads2 cases: HHHHT THHHH \(P=2*(\frac{1}{2})^5=\frac{2}{32}\). 5 consecutive heads1 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. Hello Bunuel, Can we solve this question by combination formula? Hope I am not asking wrong question. (Like for 3 consecutive heads= 5!/(3!*2!) x (1/2)^5. Using this, I am not getting 5/32. Please explain. Thanks
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Re: A fair coin is tossed 5 times. What is the probability of getting at
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25 Sep 2015, 21:31
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.
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Re: A fair coin is tossed 5 times. What is the probability of getting at
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15 May 2016, 20:31
SravnaTestPrep wrote: 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 \(8/32 = 1/4.\) How would you use this technique for atleast 2 heads ?
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