Question

A junk drawer at home contains 8 pounds fourth which work what is the probability that a randomly grab 3 pounds from the drawer and don’t end up with a Penn that works express your answer as a fraction in lowest terms or a decimal rounded to the nearest millionth

Answer 1

The container has a total of 8 pens, 4 of which work and the other 4 do not work.

You have to calculate the probability that if you choose 3 pens at random, let "N" represent the event that the pen chosen at random doesn't work, then the asked probability can be expressed as:

`P(N_1\cap N_2\cap N_3)`

The subindices 1, 2, and 3 indicate the order in which the pens are chosen.

Assuming that you do not put each pen back in the container after choosing it, you can calculate the probability of choosing each pen:

First pen: the probability that the first pen does not work is equal to the quotient of the number of pens that do not work and the total number of pens:

`\begin{gathered} P(N_1)=(4)/(8) \\ P(N_1)=(1)/(2) \end{gathered}`

After choosing the first pen, there are 7 pens remaining in the container, 3 of which do not work. These are the values you have to use to determine the next probability:

Second pen:

`P(N_2)=(3)/(7)`

After taking the second pen from the container, there are 6 pens left. Out of them, there are 2 pens that do not work. Use the remaining pens to determine the probability that the third one won't work:

Third pen:

`\begin{gathered} P(N_3)=(2)/(6) \\ P(N_3)=(1)/(3) \end{gathered}`

Finally, you can calculate the intersection between the three events by:

`P(N_1\cap N_2\cap N_3)=P(N_1)\cdot P(N_2)\cdot P(N_3)=(1)/(2)\cdot(3)/(7)\cdot(1)/(3)=(1\cdot3\cdot1)/(2\cdot7\cdot3)=(3)/(42)=(1)/(14)=0.071428`