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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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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 ?

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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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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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 [#permalink]

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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

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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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 [#permalink]

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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 ?

gmatclubot

Re: A fair coin is tossed 5 times. What is the probability of
[#permalink]
15 May 2016, 20:31

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