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

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.

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%.