Suppose that you flip a coin 13 times. What is the probability that you achieve at least 5 tails?

1 answer:

5 0

Heads, or tails. It's 1/2 a chance, because either way you will either land heads or tails. It would be a probability of 1-(1/2)^5. Can you solve that?

You might be interested in

At a Noodles & Company restaurant, the probability that a customer will order a nonalcoholic beverage is .51.

satela [25.4K]

Answer:

$P(x\geq 7)=0.1886$

Step-by-step explanation:

The variable x, that said the number of customer that will order a nonalcoholic beverage in a sample of n customers follows a binomial distribution. Because we have n identical and independent events with a probability p of success and (1-p) of fail.

So, the probability that x customers will order a nonalcoholic beverage is:

$P(x)=\frac{n!}{x!(n-x)!}*p^{x}*(1-p)^{n-x}$

Where n is the size of the sample and p is the probability that a customer order a nonalcoholic beverage, so replacing the values, we get:

$P(x)=\frac{10!}{x!(10-x)!}*0.51^{x}*(1-0.51)^{10-x}$

Now, the probability that at least 7 will order a nonalcoholic beverage is equal to:

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

Where:

$P(7)=\frac{10!}{7!(10-7)!}*0.51^{7}*(1-0.51)^{10-7}=0.1267\\P(8)=\frac{10!}{8!(10-8)!}*0.51^{8}*(1-0.51)^{10-8}=0.0494\\P(9)=\frac{10!}{9!(10-9)!}*0.51^{9}*(1-0.51)^{10-9}=0.0114\\P(10)=\frac{10!}{10!(10-10)!}*0.51^{10}*(1-0.51)^{10-10}=0.0011$