# A coin is tossed 12 times. What is the probability of getting exactly 6 tails?

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What is the probability of getting 6 consecutive tails?

What is the probability of getting 6 consecutive tails?

The probability is

Now let us consider the following:

Here,

Now, substituting this into (1), we get

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To find the probability of getting exactly 6 tails when a coin is tossed 12 times, you can use the binomial probability formula, which is:

[ P(X = k) = \binom{n}{k} \times p^k \times (1-p)^{n-k} ]

Where:

- ( n ) is the number of trials (in this case, 12 tosses).
- ( k ) is the number of successes (in this case, 6 tails).
- ( p ) is the probability of success on each trial (for a fair coin, ( p = 0.5 )).
- ( \binom{n}{k} ) represents the number of combinations of ( n ) items taken ( k ) at a time.

Substitute the values into the formula:

[ P(X = 6) = \binom{12}{6} \times (0.5)^6 \times (1-0.5)^{12-6} ]

[ P(X = 6) = \binom{12}{6} \times (0.5)^6 \times (0.5)^6 ]

Calculate:

[ P(X = 6) = \binom{12}{6} \times (0.5)^{12} ]

[ P(X = 6) = \frac{12!}{6!(12-6)!} \times 0.5^{12} ]

[ P(X = 6) = \frac{479001600}{720 \times 720} \times 0.5^{12} ]

[ P(X = 6) = \frac{479001600}{518400} \times 0.5^{12} ]

[ P(X = 6) ≈ 0.2256 ]

So, the probability of getting exactly 6 tails when a coin is tossed 12 times is approximately 0.2256.

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To find the probability of getting exactly 6 tails when a coin is tossed 12 times, you can use the binomial probability formula.

The binomial probability formula is:

[ P(X = k) = \binom{n}{k} \times p^k \times (1-p)^{n-k} ]

Where:

- ( n ) is the total number of trials (in this case, the number of coin tosses)
- ( k ) is the number of successes (in this case, the number of tails)
- ( p ) is the probability of success on each trial (for a fair coin, ( p = 0.5 ))
- ( \binom{n}{k} ) is the number of combinations of ( n ) items taken ( k ) at a time (this represents the number of ways to get exactly ( k ) tails in ( n ) tosses)

Substituting the given values into the formula:

[ P(X = 6) = \binom{12}{6} \times (0.5)^6 \times (1-0.5)^{12-6} ]

[ P(X = 6) = \binom{12}{6} \times (0.5)^6 \times (0.5)^6 ]

[ P(X = 6) = \binom{12}{6} \times (0.5)^{12} ]

Now, calculate ( \binom{12}{6} ):

[ \binom{12}{6} = \frac{12!}{6!(12-6)!} = \frac{12!}{6!6!} ]

[ = \frac{12 \times 11 \times 10 \times 9 \times 8 \times 7}{6 \times 5 \times 4 \times 3 \times 2 \times 1} ]

[ = 924 ]

Now, substitute this value into the equation:

[ P(X = 6) = 924 \times (0.5)^{12} ]

[ P(X = 6) \approx 0.2256 ]

So, the probability of getting exactly 6 tails when a coin is tossed 12 times is approximately 0.2256, or about 22.56%.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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