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The probability is 1/2 that a certain coin will turn up head [#permalink]

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26 Dec 2012, 05:53

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The probability is 1/2 that a certain coin will turn up heads on any given toss. If the coin is to be tossed three times, what is the probability that on at least one of the tosses the coin will turn up tails?

The probability is 1/2 that a certain coin will turn up heads on any given toss. If the coin is to be tossed three times, what is the probability that on at least one of the tosses the coin will turn up tails?

Re: The probability is 1/2 that a certain coin will turn up head [#permalink]

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04 Jan 2013, 17:27

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Although Bunuel's approach is, also in my opinion, the best way to go for this sort of question, you could also arrive at the same answer by using the following line of thought:

To get at least 1 tails, you can get one of the 3 configurations (in no particular order):

H H T -> 3 * 1/2*1/2*1/2 H T T -> 3 * 1/2*1/2*1/2 T T T -> 1/2*1/2*1/2

Re: The probability is 1/2 that a certain coin will turn up head [#permalink]

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16 Sep 2013, 11:09

caioguima wrote:

Although Bunuel's approach is, also in my opinion, the best way to go for this sort of question, you could also arrive at the same answer by using the following line of thought:

To get at least 1 tails, you can get one of the 3 configurations (in no particular order):

H H T -> 3 * 1/2*1/2*1/2 H T T -> 3 * 1/2*1/2*1/2 T T T -> 1/2*1/2*1/2

P = 3/8 + 3/8 + 1/8 = 7/8

Hi,

Kindly explain why it is 3* 1/2 * 1/2 * 1/2. According to my understanding, the prob of head is 1/2 and tail is 1/2. So HHT just has to be 1/2 * 1/2 * 1/2.. Isn't it? why multiply by 3? Kindly clarify.

Although Bunuel's approach is, also in my opinion, the best way to go for this sort of question, you could also arrive at the same answer by using the following line of thought:

To get at least 1 tails, you can get one of the 3 configurations (in no particular order):

H H T -> 3 * 1/2*1/2*1/2 H T T -> 3 * 1/2*1/2*1/2 T T T -> 1/2*1/2*1/2

P = 3/8 + 3/8 + 1/8 = 7/8

Hi,

Kindly explain why it is 3* 1/2 * 1/2 * 1/2. According to my understanding, the prob of head is 1/2 and tail is 1/2. So HHT just has to be 1/2 * 1/2 * 1/2.. Isn't it? why multiply by 3? Kindly clarify.

The point is that two heads and a tail can occur in three ways: HHT, HTH, THH. The probability of each case is 1/2*1/2*1/2.

Re: The probability is 1/2 that a certain coin will turn up head [#permalink]

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20 Nov 2013, 04:41

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Bunuel wrote:

Walkabout wrote:

The probability is 1/2 that a certain coin will turn up heads on any given toss. If the coin is to be tossed three times, what is the probability that on at least one of the tosses the coin will turn up tails?

Bunuel, would you mind explaining how you find the power to which you have to raise? What if e.g. there is a bag with three marbles, blue, red and yellow. Now the question is e.g. "What is the probabilty to get a blue marble on at least 1 try if you try 4 times" (putting the marbles back all the time).

Would it be P(at least 1 blue marble) = 1 - P(none blue) = 1 - (2/3)^4 ??

The probability is 1/2 that a certain coin will turn up heads on any given toss. If the coin is to be tossed three times, what is the probability that on at least one of the tosses the coin will turn up tails?

Bunuel, would you mind explaining how you find the power to which you have to raise? What if e.g. there is a bag with three marbles, blue, red and yellow. Now the question is e.g. "What is the probabilty to get a blue marble on at least 1 try if you try 4 times" (putting the marbles back all the time).

Would it be P(at least 1 blue marble) = 1 - P(none blue) = 1 - (2/3)^4 ??

Thanks for your explanation!

Yes, that's correct, the power must be the number of tries.

For the original question: P(at last 1 tails) = 1 - P(all heads) = 1 - (1/2*1/2*1/2)= 1 - (1/2)^3 = 7/8.

For your example: P(at least 1 blue) = 1 - P(no blue) = 1- (2/3*2/3*2/3*2/3) = 1- (2/3)^4.

Re: The probability is 1/2 that a certain coin will turn up head [#permalink]

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28 Nov 2014, 10:48

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First thing to do is to come up with the total number of possible outcomes: The coin is tossed 3 times and there is an equal probability that the coin will turn up heads or tail on each toss (which means that each toss has only two possible outcomes) _ _ _ 2*2*2=2^3=8 Multiply the number of possible outcomes per toss to arrive at the total number of possible outcomes. If the Question would state that the coin is to be tossed four times, the total number of possible outcomes would simply imply another multiplication by 2 or 2^4 which is 16.

Next step is to find the number of scenarios that fulfill the condition the Question stem asks for. (At least one tail) One can easily recognize that ALL scenarios BUT ONE will include at least one tail. I am talking about the scenario in which all three tosses result in heads. --> HHH So 7 out of 8 scenarios will include at least one tail. THH, HTH, HHT etc… This is already your final answer. P(at least one tail) = 7/8

Re: The probability is 1/2 that a certain coin will turn up head [#permalink]

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The probability is 1/2 that a certain coin will turn up heads on any given toss. If the coin is to be tossed three times, what is the probability that on at least one of the tosses the coin will turn up tails?

(A) 1/8 (B) 1/2 (C) 3/4 (D) 7/8 (E) 15/16

In this problem, there are only two events that could occur for the 3 coin flips. Either the coin will land on tails zero times, or the coin will land on tails at least one time. (Remember that the phrase "at least one time" means "one or more."

Writing this in a probability statement yields:

P(landing on tails at least 1 time) + P(landing on tails zero times) = 1

Thus, we can say:

P(landing on tails at least 1 time) = 1 - P(landing on tails zero times)

Since we are tossing the coin 3 times, the outcome of zero tails in 3 tosses is the same as getting heads on all 3 tosses. We can calculate the probability of zero tails in 3 tosses as the probability of 3 heads in 3 tosses:

½ x ½ x ½ =1/8

Plugging this into our formula we have:

P(landing on tails at least 1 time) = 1 – 1/8

P(landing on tails at least 1 time) = 7/8

Answer is D.
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