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# The probability is 0.6 that an “unfair” coin will turn up

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Joined: 25 Oct 2008
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The probability is 0.6 that an “unfair” coin will turn up [#permalink]

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02 Nov 2009, 04:51
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Difficulty:

35% (medium)

Question Stats:

73% (01:06) correct 27% (01:22) wrong based on 373 sessions

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The probability is 0.6 that an “unfair” coin will turn up tails on any given toss. If the coin is tossed 3 times, what is the probability that at least one of the tosses will turn up tails?

A. 0.064
B. 0.36
C. 0.64
D. 0.784
E. 0.936

Where's my error:
P(1)+P(2)+P(3)
6/10.4/10.4/10 + 6/10.6/10.4/10 +6/10.6/10.6/10
=0.516
[Reveal] Spoiler: OA

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Last edited by Bunuel on 11 Mar 2012, 13:07, edited 1 time in total.
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02 Nov 2009, 05:05
I'm getting. Someone correct me if i'm wrong

First time:
probability of getting tails is .6 = .6

2nd time:
probability of getting heads is (.4)(.6) = .24

3rd time:
probability of getting tails is (.4)(.4)(.6) = .144

Probability of flipping unfair coin and getting tails is .984

Last edited by lagomez on 02 Nov 2009, 05:17, edited 2 times in total.
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Joined: 29 Oct 2009
Posts: 208
GMAT 1: 750 Q50 V42

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02 Nov 2009, 05:09
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solve it this way:

probability that at least one of the tosses will turn up tails = 1 - probability that all will be heads

= 1 - (0.4*0.4*0.4)

= 1 - 0.064

= 0.936

where you went wrong:

probability that 1 is tails = (.6*.4*.4)+(.4*.6*.4)+(.4*.4*.6)

similarly for probability that 2 is tails.. consider all the cases..

in your solution, you have considered only one case each for both of the above..

even if you had solved it correctly, this method is longer and more complicated.. solve it as shown in my post above.. easier and quicker..

hope this helps!

cheers!
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11 Mar 2012, 12:46
required probability = 1 - p(3 heads)

$$= 1 - (\frac{2}{5})^3 = \frac{117}{125}$$
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11 Mar 2012, 13:05
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tejal777 wrote:
The probability is 0.6 that an “unfair” coin will turn up tails on any given toss. If the coin is tossed 3 times, what is the probability that at least one of the tosses will turn up tails?

Where's my error:
P(1)+P(2)+P(3)
6/10.4/10.4/10 + 6/10.6/10.4/10 +6/10.6/10.6/10
=0.516

The problem with your solution is that 1 or 2 tails can occur in several ways:

THH can occur in 3 ways: THH, HTH and HHT (basically it's # of arrangements of 3 letters THH out of whihc 2 H's are identical, which is 3!/2!=3). So, you should have P(THH)=0.6*0.4^2*3!/2!;
Similarly TTH can also occur in 3 way: TTH, THT and HTT. So, you should have P(TTH)=0.6^2*0.4*3!/2!.
Notice that we don't have the same issue with TTT, since it can occur only in one way: TTT.

$$P(T\geq{1})=0.6*0.4^2*\frac{3!}{2!}+0.6^2*0.4*\frac{3!}{2!}+0.6^3=0.936$$.

Though this question can be solved with an easier approach: $$P(T\geq{1})=1-P(H=3)=1-0.4^3=0.936$$ (the probability of at least one tails is 1 minus the probability of zero tails).

Hope it's clear.
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Re: The probability is 0.6 that an “unfair” coin will turn up [#permalink]

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08 Dec 2014, 07:07
Wy we do not multiply 0.4^3by 3!/2! ?
So as a result we will have = 1- 0.4^3*3!/2!=0.808
I do understand that this is incorrect answer, but could not explain to myself what is wrong with it.
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Joined: 02 Sep 2009
Posts: 43898
Re: The probability is 0.6 that an “unfair” coin will turn up [#permalink]

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08 Dec 2014, 07:16
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Expert's post
Van4ez wrote:
Wy we do not multiply 0.4^3by 3!/2! ?
So as a result we will have = 1- 0.4^3*3!/2!=0.808
I do understand that this is incorrect answer, but could not explain to myself what is wrong with it.

Factorial correction (multiplying by x!/(y!...)) account for different ways an event can occur. For example, THH can occur in 3 ways: THH, HTH and HHT, so we are multiplying by 3!/2! = 3 (3!/2! is the number of permutations of 3 letters THH where 2 T's are the same).

TTT case can occur only in one way Tails, Tails, Tails, so there is no need for factorial correction for this case.

Hope it's clear.

Theory on Combinations: math-combinatorics-87345.html

DS questions on Combinations: search.php?search_id=tag&tag_id=31
PS questions on Combinations: search.php?search_id=tag&tag_id=52

Tough and tricky questions on Combinations: hardest-area-questions-probability-and-combinations-101361.html

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Re: The probability is 0.6 that an “unfair” coin will turn up [#permalink]

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12 Jan 2018, 21:00
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Re: The probability is 0.6 that an “unfair” coin will turn up   [#permalink] 12 Jan 2018, 21:00
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