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What is the probability of getting exactly three heads on
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21 Apr 2014, 21:52
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What is the probability of getting exactly three heads on five flips of a fair coin? (A) 1/32 (B) 3/32 (C) 1/4 (D) 5/16 (E) 1/2 I am looking for a method better than counting...Consider if the Q had 10 coin flips or so..i.e counting would not be a feasible or practical option
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Re: What is the probability of getting exactly three heads on
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22 Apr 2014, 01:38




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Re: What is the probability of getting exactly three heads on
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21 Apr 2014, 23:59
JusTLucK04 wrote: What is the probability of getting exactly three heads on five flips of a fair coin?
(A) 1/32 (B) 3/32 (C) 1/4 (D) 5/16 (E) 1/2
I am looking for a method better than counting...Consider if the Q had 10 coin flips or so..i.e counting would not be a feasible or practical option i use this method, hope it helps 5!/3!*2!*1/2^5 = 5/16 (D) 5!/3!*2! = 5! as we flip the coin 5 times, divided by 3! as we want heads 3 times & 2! for tails 1/2^5 as probability of getting either a heads or a tails is 1/2 raise to 5 because we flip the coin 5 times please give me kudos if it helps




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Re: What is the probability of getting exactly three heads on
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03 Sep 2016, 08:01
JusTLucK04 wrote: What is the probability of getting exactly three heads on five flips of a fair coin?
(A) 1/32 (B) 3/32 (C) 1/4 (D) 5/16 (E) 1/2
I am looking for a method better than counting...Consider if the Q had 10 coin flips or so..i.e counting would not be a feasible or practical option 5 Flips of a fair coin to get = HHHTT = no. of ways this can be achieved = 5!/3!x2! = 10 Probability to get any of the above 10 arrangements (HHHTT) = (1/2)^5 = 1/32 Total probability = 1/32 x 10 = 5/16
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Re: What is the probability of getting exactly three heads on
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20 Aug 2017, 04:37
Total number of ways to get 3 heads and 2 tails ( p(h) =p(t)= 1/2 for each) =2^5=32
# of ways 3 heads can be arranged in 5 tosses= 5c3=10 10/32=5/16 ANS



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Re: What is the probability of getting exactly three heads on
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25 Aug 2017, 18:14
2 ways: 1/ 5C2 / 2^5 => there are totally 2^5 different results, but only 5C2 favor results, 1/ (1/8) * (1/4) * 5C2



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Re: What is the probability of getting exactly three heads on
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31 Jan 2018, 00:36
HHHTT in any sequence would be required Probability would be 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 5C3 = 1/32 x 10 = 5/16 (03 heads could be in any position, so 5C3)



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Re: What is the probability of getting exactly three heads on
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08 Feb 2018, 02:39
JusTLucK04 wrote: What is the probability of getting exactly three heads on five flips of a fair coin?
(A) 1/32 (B) 3/32 (C) 1/4 (D) 5/16 (E) 1/2
I am looking for a method better than counting...Consider if the Q had 10 coin flips or so..i.e counting would not be a feasible or practical option No. of ways of selecting 3 Heads = \(5C3\) Total number = \(32\) \(P = \frac{5C3}{32}\)
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Re: What is the probability of getting exactly three heads on
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22 Jun 2018, 08:21
The answer is (5c2)*((1/2)^3)*((1/2)^2). Option D or 5/16 it is !!
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Re: What is the probability of getting exactly three heads on
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25 Jun 2018, 11:58
JusTLucK04 wrote: What is the probability of getting exactly three heads on five flips of a fair coin?
(A) 1/32 (B) 3/32 (C) 1/4 (D) 5/16 (E) 1/2 When a fair coin is flipped 5 times, there are 2^5 = 32 possible outcomes. Thus, each possible outcome is equally likely, with probability of 1/32. The number of possible outcomes for getting 3 heads and 2 tails is 5!/(3! x 2!) = (5 x 4)/2 = 10. Thus, the probability of getting 3 heads and 2 tails in 5 flips is (1/32) x 10 = 10/32 = 5/16. Answer: D
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Re: What is the probability of getting exactly three heads on
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08 Oct 2018, 22:09
Hi All, We're asked for the probability of getting EXACTLY three heads on five flips of a fair coin. This question can be approached in a couple of different ways, but they all involve a bit of 'Probability math.' To start, since each coin has two possible outcomes, there are (2)(2)(2)(2)(2) = 32 possible outcomes from flipping 5 coins. To find the number of outcomes that are EXACTLY 3 heads, you can either use the Combination Formula or do some 'brute force' math and map out all of the possibilities. By choosing 3 heads from 5 tosses, we can use the Combination Formula: N!/(K!)(NK)! = 5!/(3!)(53)! = (5)(4)(3)(2)(1)/(3)(2)(1)(2)(1) = (5)(4)/(2)(1) = 10 possible ways to flip 3 heads from 5 tosses. You could also list out the options: HHHTT HHTHT HTHHT THHHT HHTTH HTHTH THHTH HTTHH THTHH TTHHH Either way, you have 10 total options that fit what we're looking for out of a total of 32 outcomes. 10/32 = 5/16 Final Answer: GMAT assassins aren't born, they're made, Rich
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