Question

Myspace versus Facebook A recent survey suggests that $85\%$ of college students have posted a profile on Facebook,$54\%$ use Myspace regularly, and $42\%$ do both. Suppose we select a college student at random and learn that the student has a profile on Facebook. Find the probability that the student uses Myspace regularly. Show your work.

Short answer

P (My space/Facebook ) = 49.41%

Step-by-step solution

Step 1. Given Information

According to a recent survey, $85\%$ of college students have a Facebook page, $54$ percent use Myspace on a regular basis, and $42\%$ use both.

Step 2. Concept used

Complement rule: $P(notA)=1-P\left(A\right)$

Conditional probability: $P(B/A)=P(AandB)/P\left(A\right)$

Step 3. Calculation

$$\begin{gathered} P\left(Facebook\right)=85\%=0.85 \\ P(Myspace)=54\%=0.54 \end{gathered}$$

Complement rule:

$$\begin{gathered} P(Facebook\cap Myspace)=42\% \\ P(notlost)=1-P\left(lost\right)=1-0.13=0.87 \end{gathered}$$

Conditional probability is defined as follows: $P(B/A)=P(AandB)/P\left(A\right)$

Applying the conditional probability definition, we get:

$P(Myspace/Facebook)=\frac{0.42}{0.85}=\frac{42}{85}\approx 0.4941=49.41\%$