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Re: Hard probability ! [#permalink]
21 Aug 2010, 12:54
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 _________________
I will give a Fight till the End
"To dream anything that you want to dream, that is the beauty of the human mind. To do anything that you want to do, that is the strength of the human will. To trust yourself, to test your limits, that is the courage to succeed." - Bernard Edmonds
A person who is afraid of Failure can never succeed -- Amneet Padda
Re: Hard probability ! [#permalink]
23 Aug 2011, 16:51
Bunuel - U rock. +1 for you.
Cheers.! _________________
----------------------------------------------------------------------------------------- What you do TODAY is important because you're exchanging a day of your life for it! -----------------------------------------------------------------------------------------
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 ?
Re: PS: Probability [#permalink]
24 Mar 2012, 13:29
Expert's post
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. _________________
Re: A fair coin is tossed 5 times. What is the probability of [#permalink]
15 Jul 2013, 06:03
1
This post received KUDOS
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 \(8/32 = 1/4.\) _________________
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. _________________
Re: A fair coin is tossed 5 times. What is the probability of [#permalink]
22 May 2015, 23:57
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 _________________
Freedom is not a gift...It is a responsibility to pass on.....
Re: A fair coin is tossed 5 times. What is the probability of [#permalink]
25 Sep 2015, 20:31
Expert's post
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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