Question

Let Em be the event that orendum is selected RS student is male and PM equals 0.47 round solution to 3 decimal place if necessary what is the probability that randomly selected Iris student is female

Answer 1

a) The probability that a randomly selected IRSC student is female is given as:

$\begin{gathered} P(F)=1-P(M) \\ \Rightarrow P(F)=1-0.47=0.53 \\ \Rightarrow P(F)=0.53 \end{gathered}$

b) The probability of both students being male is:

$\begin{gathered} P(MM)=P(M)\times P(M) \\ \Rightarrow P(MM)=0.47\times0.47=0.2209 \\ \Rightarrow P(MM)=0.221\text{ (to thr}ee\text{ decimal places)} \end{gathered}$

c) The probability of both students being female is:

$\begin{gathered} P(FF)=P(F)\times P(F) \\ \Rightarrow P(FF)=0.53\times0.53=0.2809 \\ \Rightarrow P(FF)=0.281\text{ (to thr}ee\text{ decimal places)} \end{gathered}$

d) The probability of first student being male and the second being female is:

$\begin{gathered} P(MF)=P(M)\times P(F) \\ \Rightarrow P(MF)=0.47\times0.53=0.2491 \\ \Rightarrow P(MF)=0.249\text{ (to thr}ee\text{ decimal places)} \end{gathered}$