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Troyanec [42]
3 years ago
10

What is the likelihood that a fair coin will land heads or tails?

Mathematics
2 answers:
serious [3.7K]3 years ago
8 0

<u>Answer:</u>

The correct answer option is 0.5.

<u>Step-by-step explanation:</u>

If we have a fair coin then it means that it has a 50 % chance of landing heads up and a 50% chance of landing tails up.

A fair coin is one which has two sides: head and tail.

Since their are two sides to each fair coin, so the probability of landing for each of the side should add up to 1 (which is the total probability).

Therefore, likelihood that a fair coin will land heads or tails is 0.5.

Marina CMI [18]3 years ago
6 0

Answer:

I believe it is 0.5

Step-by-step explanation:

If you flip a normal coin (called a “fair” coin in probability parlance), you normally have no way to predict whether it will come up heads or tails. Both outcomes are equally likely. There is one bit of uncertainty; the probability of a head, written p(h), is 0.5 and the probability of a tail (p(t)) is 0.5. The sum of the probabilities of all the possible outcomes adds up to 1.0, the number of bits of uncertainty we had about the outcome before the flip. Since exactly one of the four outcomes has to happen, the sum of the probabilities for the four possibilities has to be 1.0. To relate this to information theory, this is like saying there is one bit of uncertainty about which of the four outcomes will happen before each pair of coin flips. And since each combination is equally likely, the probability of each outcome is 1/4 = 0.25. Assuming the coin is fair (has the same probability of heads and tails), the chance of guessing correctly is 50%, so you'd expect half the guesses to be correct and half to be wrong. So, if we ask the subject to guess heads or tails for each of 100 coin flips, we'd expect about 50 of the guesses to be correct. Suppose a new subject walks into the lab and manages to guess heads or tails correctly for 60 out of 100 tosses. Evidence of precognition, or perhaps the subject's possessing a telekinetic power which causes the coin to land with the guessed face up? Well,…no. In all likelihood, we've observed nothing more than good luck. The probability of 60 correct guesses out of 100 is about 2.8%, which means that if we do a large number of experiments flipping 100 coins, about every 35 experiments we can expect a score of 60 or better, purely due to chance.

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Step-by-step explanation:

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Answer:

  a. f(0) = 1

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Step-by-step explanation:

The function exists at a point if it is defined there. The function is defined anywhere on the solid line and at solid dots. It is not defined at open circles. So, the function is defined everywhere except (2, 3), which has an open circle.

The open circle at (0, 4) prevents the function from being doubly-defined at x=0, since it is already defined to be 1 at x=0.

This discussion tells you ...

  f(0) = 1

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The function has a limit at a point if approaching from the left and approaching from the right have you approaching that same point.

Consider the point (1, 2). The graph is a solid line through that point. Approaching from values less than x=1, we get to the same point (1, 2) as when we approach from values greater than x=1.

Similarly, consider the point (2, 3). Approaching from values of x less than 2, we get to the same point (2, 3) as when we approach from x-values greater than 2. The limit at x=2 is 3. The only difference from the previous case is that the function is not actually defined to be that value there.

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Now consider what happens at x=0. When we approach from the left, we approach the point (0, 4). When we approach from the right, we approach the point (0, 1). These are different points. Because they are different coming from the left and from the right, we say "the limit as x→0 does not exist."

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In summary, ...

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  b) lim x → 0 does not exist

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  d) lim x → 2 = 3

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<em>Additional comment</em>

The significance of the function not being defined at a point where the limit exists, (2, 3), is that <em>the function is not continuous there</em>. This kind of discontinuity is called "removable", because we could make the function continuous at x=2 by defining f(2) = 3 (that is, "filling the hole").

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Hi,

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