# Solved (Free): Of the 10,000 students at a certain university, 7000 have Visa cards, 6000 have MasterCards

#### ByDr. Raju Chaudhari

Mar 29, 2021

Of the 10,000 students at a certain university, 7000 have Visa cards, 6000 have MasterCards, and 5000 have both. Suppose that a student is randomly selected.

a. What is the probability that the selected student has a Visa card?
b. What is the probability that the selected student has both cards?
c. Suppose you learn that the selected individual has a Visa card (was one of the 7000 with such a card). Now what is the probability that this student has both cards?
d. Are the events has a Visa card and has a MasterCard independent? Explain.
e. Answer the question posed in Part (d) if only 4200 of the students have both cards.

#### Solution

Given that

$P(\text{Visa Card}) = \frac{7000}{10000}=0.7$.

$P(\text{Master Card}) = \frac{6000}{10000}=0.6$

$P(\text{Visa and Master Card}) = \frac{5000}{10000}=0.5$

(a) The probability that the selected student has a Visa card is


\begin{aligned} P(\text{Visa Card}) &= \frac{7000}{10000}\ &=0.7. \end{aligned}

(b) The probability that the selected student has both cards is

 \begin{aligned} P(\text{Visa and Master Card}) &= \frac{5000}{10000}\\ &=0.5 \end{aligned} 

(c) The probability that selected student has both the cards given that he has Visa card is

 \begin{aligned} P(\text{Visa and Master Card}|\text{Visa Card}) &= P(\text{Visa card} \cap \text{Master Card}|\text{Visa Card})\\ &=\frac{P(\text{Visa card} \cap \text{Master Card})}{P(\text{Visa Card})}\\ &= \frac{\frac{5000}{10000}}{\frac{7000}{10000}}\\ &=\frac{5000}{7000}\\ &= 5/7=0.71429 \end{aligned} 

(d)

 \begin{aligned} P(\text{Visa Card})\times P(\text{Master Card}) &= 0.7\times 0.6=0.42\\ &\neq 0.5 = P(\text{Visa and Master Card}) \end{aligned} 

Hence, the events has a Visa card and has a MasterCard are not independent.

(e)

Given that

 \begin{aligned} P(\text{Visa and Master Card}) &=\frac{4200}{10000}\\ =0.42 \end{aligned} 

 \begin{aligned} P(\text{Visa Card})\times P(\text{Master Card}) &= 0.7\times 0.6=0.42\\ &= 0.42 = P(\text{Visa and Master Card}) \end{aligned} `

Hence, the events has a Visa card and has a MasterCard are independent.