91Ó°ÊÓ

An experiment of tossing a nickel, dime, and a quarter.

Experiments are given as tossing a nickel, dime, and a quarter. Of interest is the side the coin lands on sample space consists of all the possible outcomes.

Thus, sample space is defined as

$S=\left\{\begin{array}{l}\left(HHH\right),\left(HHT\right),\left(HTH\right),\left(HTT\right)\\ \left(THH\right),\left(THT\right),\left(TTH\right),\left(TTT\right)\end{array}\right\}$

An experiment consists of tossing a nickel, dime, and a quarter.

Experiments are given as tossing a nickel, dime, and a quarter. Of interest is the side the coin lands on sample space consists of all the possible outcomes.

Thus, sample space is defined as

$S=\left\{\begin{array}{l}\left(HHH\right),\left(HHT\right),\left(HTH\right),\left(HTT\right)\\ \left(THH\right),\left(THT\right),\left(TTH\right),\left(TTT\right)\end{array}\right\}$

Total number of possible outcomes$=8$

Let A be the event that there are at least two tails

Possible outcomes that there are at least two tails$=\left\{\right(HTT),(THT),(TTH),(TTT\left)\right\}$

Total number of outcomes that there are at least two tails$=4$

Thus, the probability that there are at least two tails is calculated as

$$\begin{gathered} P\left(A\right)=\frac{\text{Total number of outcomes that there are at least two tails}}{\text{Total number of possible outcomes}} \\ P\left(A\right)=\frac{4}{8} \\ P\left(A\right)=0.5 \end{gathered}$$

An experiment consists of tossing a nickel, dime, and a quarter.

Experiments are given as tossing a nickel, dime, and a quarter. Of interest is the side the coin lands on sample space consists of all the possible outcomes.

Thus, sample space is defined as

$S=\left\{\begin{array}{l}\left(HHH\right),\left(HHT\right),\left(HTH\right),\left(HTT\right)\\ \left(THH\right),\left(THT\right),\left(TTH\right),\left(TTT\right)\end{array}\right\}$

Total number of possible outcomes $=8$

Let A be the event that there are at least two tails

Possible outcomes that there are at least two tails $=\left\{\right(HTT),(THT),(TTH),(TTT\left)\right\}$

Also, Let B be the event that the first and second toss lands on the head.

Possible outcomes that first and second toss lands on the head $=\left\{\right(HHH),(HHT\left)\right\}$

We observe that there is no common outcome in events Aand B.

Thus, events A and B are mutually exclusive.