Question

In a school, four out of five students have calculators.If two students are picked at random,what is the probability that a) both have a calculator b) only one has a calculator

Answer 1

- The information given to us can be written mathematically as:

`\begin{gathered} P(calc)=(4)/(5) \\ P(no\text{ }calc)=1-(4)/(5)=(1)/(5) \end{gathered}`

Question A:

- If two students are chosen at random and both of them have a calculator, this means that the probability of choosing the two of them is calculated by the AND formula.

- That is,

`\begin{gathered} P(A)\text{ AND }P(B)=P(A)* P(B) \\ \text{ Given that events A and B are independent events.} \\ \\ \text{ Since the selection of the two students is random, it means that the event of choosing} \\ \text{ the two students are independent, thus, we have:} \\ \\ P(both\text{ have calculator\rparen}=(4)/(5)*(4)/(5)=(16)/(25) \end{gathered}`

Question B:

- The probability that only one of them has a calculator implies that 1 has a calculator and the other does not have a calculator.

- Thus, we have:

`\begin{gathered} P(1\text{ student with calc\rparen}* P(1student\text{ without calc\rparen} \\ =(4)/(5)*(1)/(5)=(4)/(25) \\ \\ \text{ But there are two possible scenarios:} \\ 1.\text{ we can either pick one student with a calculator first and then one with no calculator} \\ 2.\text{ We can also pick one with no calculator first, before picking one with a calculator} \\ \\ \text{ Thus, } \\ P(\text{ Only one has a calculator})=2*(4)/(25)=(8)/(25) \end{gathered}`

Final Answer

The answers are:

`\begin{gathered} Question\text{ A:} \\ (16)/(25) \\ \\ Question\text{ B:} \\ (8)/(25) \end{gathered}`