Q. 2.10 - A First Course in Probability

Question

Sixty percent of the students at a certain school wear neither a ring nor a necklace. Twenty percent wear a ring and $30$ percent wear a necklace. If one of the students is chosen randomly, what is the probability that this student is wearing

(a) a ring or a necklace?

(b) a ring and a necklace?

Step-by-Step Solution

Answer

a) The probability that the chosen student is wearing a ring or a necklace is $0.4$

b) The probability that the chosen student is wearing a ring and a necklace is $0.1$

Part (a) Step 1. Given information.

Using the given information, 

Let R be the event that a student is wearing a ring = $P\left(R\right)=0.20$

Let N be the event that a student is wearing a necklace = $P\left(N\right)=0.30$

Students wearing neither a ring nor a necklace $=P\left(R^c\cup N^c\right)=0.60$

Part (a) Step 2. Find the probability that the chosen student is wearing a ring or a necklace.

The probability that the chosen student is wearing a ring or a necklace is $P\left(R\cup N\right)$.

$P\left(R\cup N\right)=1-P{\left(R\cup N\right)}^c$

According to De-Morgan Law

$P\left(R^c\cup N^c\right)=P{\left(R\cup N\right)}^c$

$P\left(R\cup N\right)=1-0.6=0.4$

Therefore, the probability that the chosen student is wearing a ring or a necklace is $0.4$.

Part (b) Step 1. Find the probability that the chosen student is wearing a ring and a necklace.

The probability that the chosen student is wearing a ring and a necklace is $P(R\cap N)$.

$$\begin{gathered} P\left(R\cup N\right)=P\left(R\right)+P\left(N\right)-P\left(R\cap N\right) \\ \Rightarrow 0.4=0.20+0.30-P\left(R\cap N\right) \\ \Rightarrow P\left(R\cap N\right)=0.20+0.30-0.40 \\ \Rightarrow P\left(R\cap N\right)=0.1 \end{gathered}$$

Therefore, the probability that the chosen student is wearing a ring and a necklace is $0.1$.