Give the probability accurate to five significant digits The

Give the probability accurate to five significant digits.

There are three red balls, three white balls and three blue balls in an urn. Benjamin reaches in and randomly picks three of these balls. He then puts those balls back in the urn, mixes them up, and selects three more balls. He then puts those balls back in the urn, mixes them up again, and selects three balls a third time.

A) What is the probability that he selected at least one red ball in the first drawing, no red balls in the second drawing and at least one red ball in the third drawing?

B) What is the probability that he selected at least one red ball in the first drawing, at least two red balls in the second drawing and all three red balls in the third drawing?

Solution

A) let X be the event that he selects at least one red ball in the first drawing

         Y be the event that he selects no red balls in the second drawing

          Z be the event that he selects at least one red ball in the third drawing

so P[Y]=P[he selects 3 white balls or 2 white,1 blue ball or 1 white,2blue balls or 3 blue balls]

           =³C₃*³C₀/⁹C₃+³C₂*³C₁/⁹C₃+³C₁*³C₂/⁹C₃+³C₀*³C₃/⁹C₃=0.23806

so P[X]=1-P[no red balls]=1-P[Y]=1-0.23806=0.76194

Z and X are identical as the drawings are independent

so P[Z]=0.76194

so required probability=P[XYZ]=P[X]*P[Y]*P[Z] as the drawings are independent.

                                            =0.76194*0.23806*0.76194=0.13821 [answer]

B) let K be the event that at least two red balls in the second drawing

         Q be the event that all three red balls in the third drawing.

P[Q]=³C₃*³C₀*³C₀/⁹C₃=0.01190

P[K]=P[two red balls and 1 blue ball or two red balls and 1 white ball or 3 red balls]

       =³C₂*³C₀*³C₁/⁹C₃+³C₂*³C₁*³C₀/⁹C₃+³C₃*³C₀*³C₀/⁹C₃=0.22619

so required probability=P[XKQ]=P[X]*P[K]*P[Q]=0.76194*0.22619*0.01190=0.00205 [answer]