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When a certain coin is flipped, the probability that the coin will lan

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New post 20 Apr 2017, 00:45
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When a certain coin is flipped, the probability that the coin will land on head or tail is 1/2 each. If the coin is flipped 4 times, what is the probability that it will land on tail at least twice on 4 flips?

A. \(\frac{3}{8}\)
B. \(\frac{1}{16}\)
C. \(\frac{1}{2}\)
D. \(\frac{5}{8}\)
E. \(\frac{11}{16}\)
[Reveal] Spoiler: OA

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New post 20 Apr 2017, 00:54
total no. of outcomes =16
favorable outcomes =(2h2t) 4!/2!*2! =6; (3t1h) 4!/3!=4; (4t) 4!/4!=1. Total favorable outcomes =11
req. probability=11/16-> E
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Re: When a certain coin is flipped, the probability that the coin will lan [#permalink]

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New post 20 Apr 2017, 01:50
It must land on tail atleast twice = 1- land of tail at most once

HHHH = 1
THHH = 4 ways this can happen
total ways = 1+4

Number of outcomes = 2^4 =16
P(at most once) = 5/16

P( atleast twice ) = 1 - P( at most once) =1 - 5/16 = 11/16
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New post 24 Apr 2017, 01:49
==> In general, you solve probability questions using nCr combination. In other words, from TTHH, there are (4!/2!/2!)=6 possibilities, so you get (1/2)(1/2)(1/2)(1/2)*6=3/8. From TTTH, there are (4!/3!)=4 possibilities, so you get (1/2)(1/2)(1/2)(1/2)*4=1/4. From TTTT, there is only 1 possibility, so you get (1/2)(1/2)(1/2)(1/2)=1/16 and (3/8)+(1/4)+(1/16)=(6+4+1)/16=11/16.

The answer is E.
Answer: E
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Re: When a certain coin is flipped, the probability that the coin will lan   [#permalink] 24 Apr 2017, 01:49
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