Answer: p = 0.9337
Step-by-step explanation: from the question, we have that
total number of pen (n)= 15
number of pen that has never been used=9
number of pen that has been used = 15 - 9 =6
number of pen choosing on monday = 3
total number of pen choosing on tuesday=3
note that the total number of pen is constant (15) since he returned the pen back .
probability of picking a pen that has never been used on tuesday = 9/15 = 3/5
probability of not picking a pen that has never been used on tuesday = 1-3/5=2/5
probability of picking a pen that has been used on tuesday = 6/15 = 2/5
probability of not picking a pen that has not been used on tuesday= 1- 2/5= 3/5
on tuesday, 3 balls were chosen at random and we need to calculate the probability that none of them has never been used .
we know that
probability of ball that none of the 3 pen has never being used on tuesday = 1 - probability that 3 of the pens has been used on tuesday.
to calculate the probability that 3 of the pen has been used on tuesday, we use the binomial probability distribution
p(x=r) = ![nCr * p^{r} * q^{n-r}](https://tex.z-dn.net/?f=nCr%20%2A%20p%5E%7Br%7D%20%2A%20q%5E%7Bn-r%7D)
n= total number of pens=15
r = number of pen chosen on tuesday = 3
p = probability of picking a pen that has never been used on tuesday = 9/15 = 3/5
q = probability of not picking a pen that has never been used on tuesday = 1-3/5=2/5
by slotting in the parameters, we have that
p(x=3) = ![15C3 * (\frac{2}{5})^{3} * (\frac{3}{5})^{12}](https://tex.z-dn.net/?f=15C3%20%2A%20%28%5Cfrac%7B2%7D%7B5%7D%29%5E%7B3%7D%20%2A%20%28%5Cfrac%7B3%7D%7B5%7D%29%5E%7B12%7D)
p(x=3) = 455 * ![0.4^{3} * 0.6^{12}](https://tex.z-dn.net/?f=0.4%5E%7B3%7D%20%2A%200.6%5E%7B12%7D)
p(x=3) = 455 * 0.064 * 0.002176
p(x=3) = 0.0633
thus probability that 3 of the pens has been used on tuesday. = 0.0633
probability of ball that none of the 3 pen has never being used on tuesday = 1 - 0.0633 = 0.9337