If you let X be the number of customers who pay by card,

Then X~B(20,0.3)

P(more than 6 customers who pay by card)

= 1 - P(6 or less customers who pay by card)

= 1 - P(X ≤ 6)

= 1 - 0.60801 (use binomcdf (20,0.3,6) )

= 0.39199

hold on for next part

Not so sure about this, but I am thinking is

P (as stated in the question)
= P (exactly one customer in the first 4 customers pay by credit card) * P (fifth person pays by credit card) * P (at least five more customers within the next 15 people pays by credit card)

(Updated)

@eric thank you!

At least 5 of the next 15 rather. Because more than 6 means 7 and above.

Not sure if I have to divide my result by

P (more than 6 customers out of 20 pays by credit card).

I think there is no need for it though since my earlier probability already takes this into account.

@j i don’t understand what you mean

The probability is for more than 6 customers who pay by card.

So that's a minimum of 7 people who pay by card out of the 20 people.

If the 5th customer is the second to do so, then the first customer to do so must be among one of the first 4 customers.

Since we need a minimum of 7 to pay by card,

For the remaining 15 customers, there must be a minimum of 5 people who pay by card.

Eric has edited his comment

So, if I let X be a binomial distribution for the first 4 people,

X ~ B (4, 0.3)

and if I let Y be a binomial distribution for the last 15 people,

Y ~ B (15, 0.3).

As J noted, to have more than six people in the list of 20 paying by credit card, the minimum number of customers paying by credit card in the first 20 customers must be seven.

The breakdown for this is
- exactly one out of the first four customers pay by credit card
- the fifth customer MUST pay by credit card (itself with a probability of 0.3)
- at least five of the next 15 customers pay by credit card

In short, our required probability
= P (X = 1) * 0.3 * P (Y >= 5)

We need to work out what P (X = 1) and P (Y >= 5) are equal to, before inputting our answers above.

P (Y >= 5) can be computed via 1 - P (Y <= 4).

I have no graphing calculator for this.

P(X = 1) x 0.3 x P(Y ≥ 5)

= P(X = 1) x 0.3 x (1 - P(Y ≤ 4) )

= 0.4116 (use binompdf(4,0.3, 1) ) x 0.3 x (1 - 0.51549) (use binomcdf(15,0.3, 4)

= 0.4116 x 0.3 x 0.48451

= 0.0598272948

= 0.0598 (3 s.f)

Thank you so much i understand it now!

The alternative method :

X ∼ B(20,0.3)

There are 4 ways for the 1st card-using customer to be among the first 4 customers and the 2nd card-using customer to be the 5th customer :

First Second Third Fourth Fifth

Card / Cash / Cash / Cash / Card

Cash / Card / Cash / Cash / Card

Cash / Cash / Card / Cash / Card

Cash / Cash / Cash / Card / Card

That is 4 ways out of the total number of ways for each case for the first 5 people.


Now for the case X = 7, we pick 7 out of 20 to be card paying. 2 of them are among the first 5.


Then for the other 15 card paying customers, there is no restriction so we just choose 5 out of the remaining 15.

So number of ways = 4 x 15C5


For the case of X = 8, we pick 8 out of 20.

Number of ways = 4 x 15C6


So on and so forth for X = 9 all the way to X = 17

(We cannot consider X = 18, X = 19, X = 20 as that would mean the 5th customer can never be the second to do so)



Required probability

= P(X = 7) x 4(15C5)/(20C7)
+ P(X = 8) x 4(15C6)/(20C8)
+ P(X = 9) x 4(15C7)/(20C9)
+ P(X = 10) x 4(15C8)/(20C10)
+ P(X = 11) x 4(15C9)/(20C11)
+ P(X = 12) x 4(15C10)/(20C12)
+ P(X = 13) x 4(15C11)/(20C13)
+ P(X = 14) x 4(15C12)/(20C14)
+ P(X = 15) x 4(15C13)/(20C15)
+ P(X = 16) x 4(15C14)/(20C16)
+ P(X = 17) x 4(15C15)/(20C17)

= 0.02545 + 0.01818 + 0.01002 + 0.004293 + 0.001431 + 0.0003680 + ...

= 0.0598 (3 s.f)

Notice that :

As per the formula for the binomial pdf,

P(X = 7) x 4(15C5)/(20C7)

= (20C7) x 0.3^7 x 0.7^13 x 4(15C5)/(20C7)

= 0.3^7 x 0.7^13 x (4C1)(15C5)

= (4C1)(0.3 x 0.7³) x 0.3 x (15C5)(0.3^5 x 0.7^10)

And this is actually equal to P(X=1) x 0.3 x P(Y=5) as shown in Eric's method of splitting up into 2 distributions.


Likewise,

P(X = 8) x 4(15C6)/(20C8)

= (20C8) x 0.3^8 x 0.7^12 x 4(15C6)/(20C8)

= 0.3^8 x 0.7^12 x (4C1)(15C6)

= (4C1)(0.3 x 0.7³) x 0.3 x (15C6)(0.3^6 x 0.7^9)

= P(X=1) x 0.3 x P(Y=6)


So the total probability

= P(X = 1) x 0.3 x [ P(Y = 5) + P(Y=6) + ... + P(Y = 15) ]

= P(X = 1) x 0.3 x P(Y ≥ 5)