Question

The Magazine Mass Marketing Company has received 14 entries in its latest sweepstakes. They know that the probability of receiving a magazine subscription order with an entry form is 0.4. What is the probability that no more than 3 of the entry forms will include an order

Answer 1

The probability that no more than 3 of the entry forms will include an order is P=0.1243.

Explanation:

We can model the probability of having k entries with order included within the 14 entries with a binomial distribution, with parameters n=14 and p=0.4.

The probabiltity of having k orders in the sample is:

`P(x=k) = \dbinom{n}{k} p^(k)(1-p)^(n-k)\\\\\\P(x=k) = \dbinom{14}{0} 0.4^(k) 0.6^(14-k)\\\\\\`

We have to calculate the probability that no more than 3 of the entry forms will include an order. This is P(X≤3).

`P(X\leq3)=P(x=0)+P(x=1)+P(X=2)+P(x=3)\\\\\\P(x=0) = \dbinom{14}{0} p^(0)(1-p)^(14)=1*1*0.0008=0.0008\\\\\\P(x=1) = \dbinom{14}{1} p^(1)(1-p)^(13)=14*0.4*0.0013=0.0073\\\\\\P(x=2) = \dbinom{14}{2} p^(2)(1-p)^(12)=91*0.16*0.0022=0.0317\\\\\\P(x=3) = \dbinom{14}{3} p^(3)(1-p)^(11)=364*0.064*0.0036=0.0845\\\\\\\\ P(X\leq3)=0.0008+0.0073+0.0317+0.0845=0.1243`