Question

Set up an appropriate sample space for each of Problems 1.1 to 1.10 and use itto solve the problem. Use either a uniform or non-uniform sample space or try both.

Three coins are tossed; what is the probability that two are heads and one tails? That the first two are heads and the third tails? If at least two are heads, what is the probability that all are heads?

Short answer

The required sample space is .$hhh,hht,hth,htt,thh,tht,tth,ttt$

The probability that two heads and a tail is.$\frac{3}{8}$

The probability that the first two are heads and third one is tail is.$\frac{1}{8}$

The probability that when at least two are heads, all are heads is.$\frac{1}{4}$

Step-by-step solution

Definition of Sample Space

Sample space of any experiment is the set of all possible mutually exclusive events or experiments. For instance, if a coin is tossed, then the possible outcome is head and tail.

Creation of the sample space

When three coins are tossed, with each toss there is a possibility of getting head or tail. The sample space is as follows,

$hhh,hht,hth,htt,thh,tht,tth,ttt$

It is observed that the sample space contains 8 points. Each point of the obtained sample space has an equal probability of $\frac{1}{8}$.

Determination of the probability that two are heads and one tails

It can be observed that two heads and a tail are 3 that are,$hht,hth,thh$ .

Find the probability that the two are heads and one tails by adding the probabilities of each possible outcomes.

$\begin{array}{c}p=\frac{1}{8}+\frac{1}{8}+\frac{1}{8}\\ =\frac{3}{8}\end{array}$

Thus, the probability that two heads and a tail is.$\frac{3}{8}$

Determination of the probability that first two are heads and third is tails

From the obtained sample space, it can be observed that first two heads and a tail is 1 that is, $hht$.

Find the probability that the first two are heads and third one is tail.

$p=\frac{1}{8}$

Thus, the probability that the first two are heads and third one istail is.$\frac{1}{8}$

Determination of the probability that when at least two are heads, all are heads

From the obtained sample space, it can be observed that the outcomes with at least two heads are 4that are,, and each point of the sample space has probability $\frac{1}{4}$.

Find the probability that when at least two are heads, all are heads.$p=\frac{1}{4}$

Thus, the probability that when at least two are heads, all are heads is.$\frac{1}{4}$