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Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 7612
GMAT 1: 760 Q51 V42 GPA: 3.82
If a certain coin is flipped, it has probability ½ of landing on heads  [#permalink]

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Difficulty:   5% (low)

Question Stats: 83% (00:59) correct 17% (01:33) wrong based on 63 sessions

### HideShow timer Statistics [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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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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Intern  B
Joined: 03 Nov 2018
Posts: 3
Re: If a certain coin is flipped, it has probability ½ of landing on heads  [#permalink]

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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
e-GMAT Representative V
Joined: 04 Jan 2015
Posts: 2943
Re: If a certain coin is flipped, it has probability ½ of landing on heads  [#permalink]

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3

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.

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Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 7612
GMAT 1: 760 Q51 V42 GPA: 3.82
Re: If a certain coin is flipped, it has probability ½ of landing on heads  [#permalink]

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=>

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

_________________ 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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