Question

A coin is flipped 15 times where each flip comes up either heads or tails. How many possible outcomes (a) contain exactly four tails (b) contain at least three heads?

Answer 1

Using the arrangements formula, it is found that:

a) There are 1365 possible outcomes that contain exactly four tails.

b) There are 1,307,674,399,889 possible outcomes that contain at least three heads.

What is the arrangements formula?

The number of possible arrangements of n elements is given by the factorial of n, that is:

$A_n = n!$

When there are repetitions, the number of ways is given as follows:

$A_{n}^{n_1, n_2, \cdots, n_n} = \frac{n!}{n_1!n_2! \cdots n_n!}$

In which $n_1, n_2, \cdots, n_n$ are the numbers of repetitions.

Item a:

11 heads and 4 tails, hence:

$A_{15}^{11,4} = \frac{15!}{11!4!} = 1365$

There are 1365 possible outcomes that contain exactly four tails.

Item b:

The total number is:

$A_{15} = 15! = 1,307,674,400,000$

With no heads:

$A_{15}^{15,0} = \frac{15!}{15!0!} = 1$

With one head:

$A_{15}^{14,1} = \frac{15!}{14!1!} = 15$

With two heads:

$A_{15}^{13,2} = \frac{15!}{13!2!} = 95$

Hence the number of outcomes with less than three heads is:

1 + 15 + 95 = 111

With at least three heads, the number of outcomes is:

$1,307,674,400,000 - 111 = 1,307,674,399,889$

There are 1,307,674,399,889 possible outcomes that contain at least three heads.

More can be learned about the arrangements formula at https://brainly.com/question/24648661

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