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# Ten coins are tossed simultaneously. In how many of the outcomes will

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Math Expert
Joined: 02 Sep 2009
Posts: 43314

Kudos [?]: 139342 [0], given: 12786

Ten coins are tossed simultaneously. In how many of the outcomes will [#permalink]

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14 Mar 2016, 07:31
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45% (medium)

Question Stats:

64% (00:58) correct 36% (01:04) wrong based on 105 sessions

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Ten coins are tossed simultaneously. In how many of the outcomes will the third coin turn up a head?

A. 2^9
B. 2^10
C. 3 * 2^8
D. 3 * 2^9
E. 3 * 2^10
[Reveal] Spoiler: OA

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Kudos [?]: 139342 [0], given: 12786

Manager
Joined: 09 Jun 2015
Posts: 102

Kudos [?]: 9 [0], given: 0

Ten coins are tossed simultaneously. In how many of the outcomes will [#permalink]

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14 Mar 2016, 07:40
Bunuel wrote:
Ten coins are tossed simultaneously. In how many of the outcomes will the third coin turn up a head?

A. 2^9
B. 2^10
C. 3 * 2^8
D. 3 * 2^9
E. 3 * 2^10

Fix the third coin as H. The remaining 9 coins have 2^9 outcomes.

Kudos [?]: 9 [0], given: 0

Intern
Joined: 16 Apr 2015
Posts: 35

Kudos [?]: 38 [1], given: 124

Re: Ten coins are tossed simultaneously. In how many of the outcomes will [#permalink]

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14 Mar 2016, 11:23
1
KUDOS
Bunuel wrote:
Ten coins are tossed simultaneously. In how many of the outcomes will the third coin turn up a head?

A. 2^9
B. 2^10
C. 3 * 2^8
D. 3 * 2^9
E. 3 * 2^10

Please note this question is about outcome not arrangement, one of most common areas of confusion.
Apart from 3rd toss which needs to be Heads all other outcome are free and independent of each other.

total outcome for remaining 9 toss = $$2^9$$
3rd toss out come => 1 way

hence total outcome =>$$2^9 * 1 = 2^9$$=> A

Kudos [?]: 38 [1], given: 124

Math Expert
Joined: 02 Aug 2009
Posts: 5522

Kudos [?]: 6418 [0], given: 122

Ten coins are tossed simultaneously. In how many of the outcomes will [#permalink]

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04 Nov 2017, 19:49
Expert's post
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Bunuel wrote:
Ten coins are tossed simultaneously. In how many of the outcomes will the third coin turn up a head?

A. 2^9
B. 2^10
C. 3 * 2^8
D. 3 * 2^9
E. 3 * 2^10

Hi...
FIRST point :- If there are no restrictions, outcomes =2*2*2...10times=2^10
each coin can have two outcomes : head or tail, so 2*2*2...
SECOND point :- the restrictions, third to be HEAD only...
Since third coin has 2 outcomes, half of total will have head and half tail..
So outcomes = $$\frac{2^{10}}{2}=2^9$$

A
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Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

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Kudos [?]: 6418 [0], given: 122

Manager
Joined: 24 Jun 2017
Posts: 118

Kudos [?]: 16 [0], given: 129

Re: Ten coins are tossed simultaneously. In how many of the outcomes will [#permalink]

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09 Nov 2017, 21:21
law of large numbers:) or symmetry in case of permutations
50% of total outcomes - 2^10/2 = 2^9

Kudos [?]: 16 [0], given: 129

Target Test Prep Representative
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 1820

Kudos [?]: 1046 [2], given: 5

Re: Ten coins are tossed simultaneously. In how many of the outcomes will [#permalink]

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14 Nov 2017, 06:23
2
KUDOS
Expert's post
Bunuel wrote:
Ten coins are tossed simultaneously. In how many of the outcomes will the third coin turn up a head?

A. 2^9
B. 2^10
C. 3 * 2^8
D. 3 * 2^9
E. 3 * 2^10

Since the total number of outcomes of the 10 coins is 2^10, and half of these outcomes will have the third coin turning up as a head (the other half will have the third coin as a tail), we see that the number of outcomes with the third coin turning up as a head is ½ x 2^10 = 2^9.

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Jeffery Miller

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Kudos [?]: 1046 [2], given: 5

Re: Ten coins are tossed simultaneously. In how many of the outcomes will   [#permalink] 14 Nov 2017, 06:23
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