Step-by-step explanation:
Ok, so we can use the binomial distribution formula, since certain constraints are met in this case:
- each event is independent
- there is either a failure or a success
- there is a set number of trials
- the probability of success is constant (basically independence)
<u><em>Binomial Distribution Formula:</em></u>
![P(x)=(^n_x)*p^{x}*(1-p)^{n-x}](https://tex.z-dn.net/?f=P%28x%29%3D%28%5En_x%29%2Ap%5E%7Bx%7D%2A%281-p%29%5E%7Bn-x%7D)
In this formula n=number of trials
p = probability of success
(1-p) = probability of failure
x = number of success
and P(x) = probability of getting "x" successes
The combination formula can be calculated using the following formula: ![(^n_k) =\frac{n!}{k!(n-k)!}](https://tex.z-dn.net/?f=%28%5En_k%29%20%3D%5Cfrac%7Bn%21%7D%7Bk%21%28n-k%29%21%7D)
So let's define the values in the equation:
The x depends on the part of the question we're on, so let's start with part A:
- So in this case x=0, since the number of "successes" in this context just means how many adults regret getting a tattoo.
- So plugging in the values into the binomial distribution formula we get the following equation:
![P(0)=\frac{8!}{0!*(8-0)!}*0.28^0*0.72^{8-0}](https://tex.z-dn.net/?f=P%280%29%3D%5Cfrac%7B8%21%7D%7B0%21%2A%288-0%29%21%7D%2A0.28%5E0%2A0.72%5E%7B8-0%7D)
![P(0)=\frac{8!}{1*8!}*1*0.72^8](https://tex.z-dn.net/?f=P%280%29%3D%5Cfrac%7B8%21%7D%7B1%2A8%21%7D%2A1%2A0.72%5E8)
- Simplifying further we get:
![P(0)=0.72^8](https://tex.z-dn.net/?f=P%280%29%3D0.72%5E8)
- Now just use a calculator to calculate this value:
![P(0)\approx0.0722](https://tex.z-dn.net/?f=P%280%29%5Capprox0.0722)
- This can also be represented as 7.22%, by multiplying this by 100
Part B:
So in this case x=1, so plugging this into the formula we get:
![P(1)=\frac{8!}{1!*(8-1)!}*0.28^1*0.72^{8-1}\\\\P(1)=\frac{8*7!}{7!}*0.28*0.72^7\\P(1)\approx 8*0.28*0.1003\\P(1)\approx 0.2247](https://tex.z-dn.net/?f=P%281%29%3D%5Cfrac%7B8%21%7D%7B1%21%2A%288-1%29%21%7D%2A0.28%5E1%2A0.72%5E%7B8-1%7D%5C%5C%5C%5CP%281%29%3D%5Cfrac%7B8%2A7%21%7D%7B7%21%7D%2A0.28%2A0.72%5E7%5C%5CP%281%29%5Capprox%208%2A0.28%2A0.1003%5C%5CP%281%29%5Capprox%200.2247)
This can also be represented as 22.47%
Part C:
So in this case, all we really have to do is add the probabilities, as you may recall the formula: ![P(A\cup B)=P(A)+P(B)-P(A\cap B)](https://tex.z-dn.net/?f=P%28A%5Ccup%20B%29%3DP%28A%29%2BP%28B%29-P%28A%5Ccap%20B%29)
In this case, it's mutually exclusive, because there is no chance of getting 0 and 1 at the same time, so the probability of P(A and B) is just 0, so we simplify the equation to: ![P(A\cup B)=P(A)+P(B)](https://tex.z-dn.net/?f=P%28A%5Ccup%20B%29%3DP%28A%29%2BP%28B%29)
So adding the two probabilities calculated in the previous questions we get: ![P(x\le1)\approx 0.2247+0.0722\\P(x\le1) \approx 0.2969](https://tex.z-dn.net/?f=P%28x%5Cle1%29%5Capprox%200.2247%2B0.0722%5C%5CP%28x%5Cle1%29%20%5Capprox%200.2969)
The probability can also be represented as 29.69%
Part D:
As calculated in one of the previous parts, the probability is approximately 22%, which isn't significantly low and has around a 1/5 chance of occurring so if this were to occur, there is no reason to suspect that the probabilities are incorrect.