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If a certain coin is flipped, it has probability ½ of landing on heads

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If a certain coin is flipped, it has probability ½ of landing on heads  [#permalink]

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New post 26 Dec 2018, 01:47
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[Math Revolution GMAT math practice question]

If a certain coin is flipped, it has probability \(\frac{1}{2}\) of landing on heads and probability \(\frac{1}{2}\) of landing on tails. If the coin is flipped \(4\) times, what is the probability that it will land on tails at least once?

\(A. \frac{3}{8}\)
\(B. \frac{1}{4}\)
\(C. \frac{1}{2}\)
\(D. \frac{5}{16}\)
\(E. \frac{15}{16}\)

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Re: If a certain coin is flipped, it has probability ½ of landing on heads  [#permalink]

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New post 26 Dec 2018, 02:24
MathRevolution wrote:
[Math Revolution GMAT math practice question]

If a certain coin is flipped, it has probability \(\frac{1}{2}\) of landing on heads and probability \(\frac{1}{2}\) of landing on tails. If the coin is flipped \(4\) times, what is the probability that it will land on tails at least once?

\(A. \frac{3}{8}\)
\(B. \frac{1}{4}\)
\(C. \frac{1}{2}\)
\(D. \frac{5}{16}\)
\(E. \frac{15}{16}\)



P ( of only heads) = 1/2
and 4 times would be (1/2)^4 = 1/16
P of at least one tails 1-1/16 = 15/16 IMO E
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Re: If a certain coin is flipped, it has probability ½ of landing on heads  [#permalink]

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New post 26 Dec 2018, 03:20
Probability of atleast one formula = 1- Probability of opposite event

In this case
Probability of only head = 1/2
Probability of only head when coin is flipped 4 times = (1/2)^4
= 1/16
Probabilty of atleast one tails = 1- 1/16
= 15/16
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Re: If a certain coin is flipped, it has probability ½ of landing on heads  [#permalink]

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New post 26 Dec 2018, 09:06
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Solution



Given:
When a coin is tossed,
    • Probability of head = \(\frac{1}{2}\)
    • Probability of tail = \(\frac{1}{2}\)

To find:
    • The probability of getting at least one tail, if the coin is tossed 4 times

Approach and Working:
Method 1:
P (at least 1 tail) = P (1 tail AND 3 heads) OR P (2 tails AND 2 heads) OR P (3 tails AND 1 head) OR P (4 tails) = \(\frac{1}{2}^4 * \frac{4!}{3!}\) + \(\frac{1}{2}^4 * \frac{4!}{2!2!}\) + \(\frac{1}{2}^4 * \frac{4!}{3!}\) + \(\frac{1}{2}^4\) = \(\frac{15}{16}\)

Method 2:
P (at least 1 tail) = 1 – P (all heads) = 1 – \(\frac{1}{2}^4\) = 1 – \(\frac{1}{16}\) = \(\frac{15}{16}\)

Hence, the correct answer is option E.

Answer: E

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Re: If a certain coin is flipped, it has probability ½ of landing on heads  [#permalink]

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New post 28 Dec 2018, 03:19
=>

The complementary case is the event with all heads, which has probability \((\frac{1}{2})^4\).
The probability of at least one tail is \(1 – (\frac{1}{2})^4 = 1 – \frac{1}{16} = \frac{15}{16}.\)

Therefore, the answer is E.
Answer: E
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Re: If a certain coin is flipped, it has probability ½ of landing on heads   [#permalink] 28 Dec 2018, 03:19
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