According to the manufacturer, about 31% of chewy candy in a package of cherry's chews are strawberry. what is the probability that the first strawberry candy chosen from a bag will be, at least, the fourth candy chosen overall?

0.3285

Step-by-step explanation:

This problem involves Geometric probability distribution because experiment is repeated different number of times until success is achieved.

we have to find P(X≥4). It can be written as P(X≥4)=1-P(X<4)=1-P(X≤3)

P(X≤3)=?

Using Geometric probability distribution

P(X=n)=p^x*q^n-1

P(X≤3)=P(X=1)+P(X=2)+P(X=3)

Here p=0.31,

P(X≤3)=0.31^1*0.69^0+0.31^2*0.69^1+0.31^3*0.69^2)=0.6715

P(X≥4)=1-P(X≤3)=1-0.6715=0.3285

The probability that the first strawberry candy chosen from a bag will be, at LEAST, the fourth candy chosen overall is 32.85%.

if you are saying purchasing 3 social media ads and one newspaper ad counts as one purchase then there are:

6 purchases

if you are asking how many of each then:

18 social media ads

6 newspaper ads