It is asserted that 80 percent of the cars approaching an individual toll both in New Jersey are equipped with an E-ZPass transponder. Find the probability that in a sample of six cars:

(a) All six will have the transponder. (Round your answer to 4 decimal places.)

Probability

(b) At least three will have the transponder. (Round your answer to 4 decimal places.)

Probability

(c) None will have a transponder. (Round your answer to 6 decimal places.)

Probability

### 2 Answers

(a) All six will have the transponder.

The probability of any given car having the transponder is 80% = 0.80.

The probability of all six having the transponder is the product of the individual probabilities:

0.80^6 = 0.262144

= 0.2621 (rounded as requested)

[Ambiguity here: the original probability given was a percentage. Should this have been expressed as

26.2144% “rounded to 4 decimal places”?

Combining percentage notation with specified rounding to decimal places isn’t clear.]

Let’s skip (b) and come back to it later.

(c) None will have a transponder.

The probability of any given car not having a transponder is 1 – 0.80 = 0.20.

The probability of all six not having a transponder is the product of the individual probabilities:

0.20^6 = 0.000064

(No need to round. We’ve got six decimal places.)

Now back to

b) At least three will have the transponder.

This equals the probability that none, or only one or two cars, have the transponder.

We know the probability that none do = 0.000064.

Suppose we put the cars in some particular order (say, the order in which they arrive).

Then if just one has the transponder, the probability of that particular sequence of cars having or not having it will be

0.80 (0.20^5) = 0.000256

But there are C(6,1) = 6 possible sequences like this (because the car with the transponder could be first, second, third, etc.) and each is equally probable, so the total probability of all such sequences is

C(6,1) 0.80 (0.20^5)

= 6 (0.000256) = 0.001536

Similarly, if just two of the six cars have the transponder, the number of sequences with that result is

C(6,2) = 15

and the probability of each sequence is

(0.80^2) (0.20)^4 = 0.001024

so the total probability of all such sequences is

C(6,2) (0.80^2) (0.20)^4

= 15 (0.001024) = 0.01536

With these figure, we know how to calculate the probability that at least three cars have the transponder:

1 – [0.20^6 + C(6,1) 0.80 (0.20^5) + C(6,2) (0.80^2) (0.20)^4]

= 1 – [0.000064 + 0.001536 + 0.01536]

= 1 – 0.016960

= 0.983040

= 0.9830 rounded to 4 decimal places.

(or should that have been 98.3040% “rounded to 4 decimal places”?)

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