Question

Two coins are tossed. Assume that each event is equally likely to occur. ​a) Use the counting principle to determine the number of sample points in the sample space. ​b) Construct a tree diagram and list the sample space. ​c) Determine the probability that no tails are tossed. ​d) Determine the probability that exactly one tail is tossed. ​e) Determine the probability that two tails are tossed. ​f) Determine the probability that at least one tail is tossed.

Answer 1

Answer:

(a) 4 sample points

(b) See attachment for tree diagram

(c) The probability that no tail is appeared is 1/4

(d) The probability that exactly 1 tail is appeared is 1/2

(e) The probability that 2 tails are appeared is 1/4

(f) The probability that at least 1 tail appeared is 3/4

Step-by-step explanation:

Given

$Coins = 2$

Solving (a): Counting principle to determine the number of sample points

We have:

$Coin\ 1 = \{H,T\}$

$Coin\ 2 = \{H,T\}$

To determine the sample space using counting principle, we simply pick one outcome in each coin. So, the sample space (S) is:

$S = \{HH,HT,TH,TT\}$

The number of sample points is:

$n(S) = 4$

Solving (b): The tree diagram

See attachment for tree diagram

From the tree diagram, the sample space is:

$S = \{HH,HT,TH,TT\}$

Solving (c): Probability that no tail is appeared

This implies that:

$P(T = 0)$

From the sample points, we have:

$n(T = 0) = 1$ --- i.e. 1 occurrence where no tail is appeared

So, the probability is:

$P(T = 0) = \frac{n(T = 0)}{n(S)}$

This gives:

$P(T = 0) = \frac{1}{4}$

Solving (d): Probability that exactly 1 tail is appeared

This implies that:

$P(T = 1)$

From the sample points, we have:

$n(T = 1) = 2$ --- i.e. 2 occurrences where exactly 1 tail appeared

So, the probability is:

$P(T = 1) = \frac{n(T = 1)}{n(S)}$

This gives:

$P(T = 1) = \frac{2}{4}$

$P(T = 1) = \frac{1}{2}$

Solving (e): Probability that 2 tails appeared

This implies that:

$P(T = 2)$

From the sample points, we have:

$n(T = 2) = 1$ --- i.e. 1 occurrences where 2 tails appeared

So, the probability is:

$P(T = 2) = \frac{n(T = 2)}{n(S)}$

This gives:

$P(T = 2) = \frac{1}{4}$

Solving (f): Probability that at least 1 tail appeared

This implies that:

$P(T \ge 1)$

In (c), we have:

$P(T = 0) = \frac{1}{4}$

Using the complement rule, we have:

$P(T \ge 1) + P(T = 0) = 1$

Rewrite as:

$P(T \ge 1) = 1-P(T = 0)$

Substitute known value

$P(T \ge 1) = 1-\frac{1}{4}$

Take LCM

$P(T \ge 1) = \frac{4-1}{4}$

$P(T \ge 1) = \frac{3}{4}$