Question

(1 point) You flip a fair coin 10 times. What is the probability that it lands on heads exactly 7 times? Round to four decimal places. The probability of exactly 7 heads is . What is the probability that it lands on heads at least 7 times? The probability of at least 7 heads is

Answer 1

Answer:

The probability of exactly 7 heads is 0.1172.

The probability of at least 7 heads is 0.1710

Step-by-step explanation:

For each time that the coin is tossed, there are only two possible outcomes. Either it is heads, or it is not. The probability of a toss resulting in heads is independent from other tosses. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

You flip a fair coin 10 times.

10 times, which means that $n = 10$

Fair coin means that heads and tails are equally as likely, so $p = 0.5$

What is the probability that it lands on heads exactly 7 times?

This is P(X = 7).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 7) = C_{10,7}.(0.5)^{7}.(0.5)^{3} = 0.1172$

0.1172 = 11.72% probability that it lands on heads exactly 7 times.

What is the probability that it lands on heads at least 7 times?

$P(X \geq 7) = P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 7) = C_{10,7}.(0.5)^{7}.(0.5)^{3} = 0.1172$

$P(X = 8) = C_{10,8}.(0.5)^{8}.(0.5)^{2} = 0.0439$

$P(X = 9) = C_{10,9}.(0.5)^{9}.(0.5)^{1} = 0.0098$

$P(X = 10) = C_{10,10}.(0.5)^{10}.(0.5)^{0} = 0.0001$

$P(X \geq 7) = P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.1172 + 0.0439 + 0.0098 + 0.0001 = 0.171$

0.1710 = 17.10% probability that it lands on heads at least 7 times.