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When six fair coins are tossed simultaneously, in how many of the outc

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Math Expert
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Joined: 02 Sep 2009
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When six fair coins are tossed simultaneously, in how many of the outc  [#permalink]

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New post 29 Jan 2019, 07:54
00:00
A
B
C
D
E

Difficulty:

  65% (hard)

Question Stats:

46% (02:16) correct 54% (02:05) wrong based on 43 sessions

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Intern
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Joined: 29 Jan 2019
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Re: When six fair coins are tossed simultaneously, in how many of the outc  [#permalink]

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New post 29 Jan 2019, 08:53
64 total outcomes (2^6).

0H6T - 1 possible outcome
1H5T - 6 possible outcomes
2H4T - 15 possible outcomes (5+4+3+2+1)
3H3T - screw figuring it out, it has to be symmetrical with tails anyway so I'll come back to it

4H2T - 15 outcomes
5H1T - 6 outcomes
6H0T - 1 outcome

3H3T = 64 - (15+6+1) - (15+6+1) = 20

1+6+15+20 = 42.

E is the answer.
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Re: When six fair coins are tossed simultaneously, in how many of the outc  [#permalink]

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New post 29 Jan 2019, 10:07
Bunuel wrote:
When six fair coins are tossed simultaneously, in how many of the outcomes will at most three of the coins turn up as heads?

A. 22

B. 31

C. 32

D. 41

E. 42


at most 3 means ; to the max 3
we will have 4 cases:
case1 ; flip coin 6 times we get 0 H and 6 T ; 6c0
case2: 1 H and 5 T ; 6c1
case3 ; 2 H and 4 T= 6c2
case 4 : 3 H and 3 T = 6c3

so total outcomes to get H at most 3 times
6c0+6c1+6c2+6c3 = 1+6+15+20 = 42
IMO E
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Re: When six fair coins are tossed simultaneously, in how many of the outc  [#permalink]

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New post 01 Feb 2019, 08:06
Bunuel wrote:
When six fair coins are tossed simultaneously, in how many of the outcomes will at most three of the coins turn up as heads?

A. 22

B. 31

C. 32

D. 41

E. 42


#(at most 3 heads) = #(0 head) + #(1 head) + #(2 heads) + #(3 heads)
= 6C0 + 6C1 + 6C2 + 6C3

= 1 + 6 + (6 x 5)/2 + (6 x 5 x 4)/(3 x 2)

= 1 + 6 + 15 + 20

= 42

Answer: E
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Re: When six fair coins are tossed simultaneously, in how many of the outc   [#permalink] 01 Feb 2019, 08:06
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