Question

-3 points Suppose we have a fair coin that is equally likely to come up heads or tails (a) The coin is flipped 3 times. How many ways can we get exactly 1 head? (b) The coin is flipped 5 times. How many ways can we get exactly 2 tails? (c) The coin is flipped 4 times. How many ways can we get at least 3 tails?

Answer 1

Solution:

We are given that: a fair coin is flipped in following ways:

Part a) The coin is flipped 3 times.

We have to find number of ways we can get exactly 1 head.

We use combination formula:

$\dpi{100} _{n}C_{x}=\frac{n!}{x!\times (n-x)!}$

n = Number of times coin is flipped = 3

x = number of times we get Head = 1

thus

$\dpi{100} _{3}C_{1}=\frac{3!}{1!\times (3-1)!} =\frac{3!}{1! \times 2!}=\frac{3 \times 2 \times 1}{1 \times 2 \times 1}=3$

Thus the number of ways we can get exactly 1 head = 3

Part b) The coin is flipped 5 times.

We have to find number of ways we can get exactly 2 tails.

Thus here we use

n = Number of times coin is flipped = 5

x = number of times we get Tails= 2

Thus

$\dpi{100} _{5}C_{2}=\frac{5!}{2!\times (5-2)!} =\frac{5!}{2! \times 3!}=\frac{5 \times 4\times 3 \times 2 \times 1}{2 \times 1 \times 3 \times 2 \times 1}=10$

Thus the number of ways we can get exactly 2 tails = 10

Part c) The coin is flipped 4 times.

We have to find number of ways we can get at least 3 tails.

At least 3 tails means $\dpi{100} x \geq 3$

thus possible values of x are 3 and 4

means number of tails are 3 or 4.

Lets consider x = 3 and n = number of times coin flipped = 4

thus

$\dpi{100} _{4}C_{3}=\frac{4!}{3!\times (4-3)!} =\frac{4!}{3! \times 1!}=\frac{4\times 3 \times 2 \times 1}{ 3 \times 2 \times 1 \times 1}=4$

and now consider x = 4 and n = 4

$\dpi{100} _{4}C_{4}=\frac{4!}{4!\times (4-4)!} =\frac{4!}{4! \times 0!}=\frac{4\times 3 \times 2 \times 1}{ 4 \times 3 \times 2 \times 1 \times 1}=1$

Total number of ways in which we can get at least 3 tails = 4 + 1 = 5