Question

Suppose there are 100 students in your accounting class, 10 of whom are left-handed. Two students are selected at random.

a) Draw a probability tree and insert the probabilities for each branch.

b) What is the probability of the following events?

c) Both are right-handed.

d) Both are left-handed.

e) One is right-handed and the other is left-handed.

f) At least one is right-handed

Answer 1

In accounting class there are 10 left-handed students out of 100 students.

Approximately \(10 \%\) of people are left-handed. So, the probability of left-handed people is

$$ \begin{aligned} P(\text { Left-handed }) &=10 \% \\ &=0.1 \\ P(\text { Right-handed }) &=1-0.1 \\ &=0.9 \end{aligned} $$

(a) We draw a probability tree for the above experiment and also we assign the probabilities to each branch.

(b) If two people are selected at random, then we find the probability of both people is righthanded,

$$ \begin{aligned} P(\text { both are right-handed }) &=0.9 \times 0.9[\text { independent events }] \\ &=0.81 \end{aligned} $$

(c) If two people are selected at random, then we find the probability of both people is lefthanded,

$$ \begin{aligned} P(\text { both are left-handed }) &=0.1 \times 0.1 \quad[\text { independent events }] \\ &=0.01 \end{aligned} $$

(d) If two people are selected at random, then we find the probability of one is right-handed and the other is left-handed,

\(P(\) one is left-handed and other is right-handed \()=0.1 \times 0.9+0.9 \times 0.1\)

$$ \begin{array}{l} =0.9+0.9 \\ =0.18 \end{array} $$

(e) If two people are selected at random, then we find the probability at least one is right-handed,

$$ \begin{aligned} P(\text { at least one is right-handed }) &=1-P(\text { none is right-handed }) \\ &=1-P(\text { both are left-handed }) \\ &=1-0.01 \\ &=0.99 \end{aligned} $$