MATH SOLVE

4 months ago

Q:
# You have a large jar that initially contains 30 red marbles and 20 blue marbles. We also have a large supply of extra marbles of each color. Draw a marble out of the jar. If it's red, put it back in the jar, and add three red marbles to the jar from the supply of extras. If it's blue, put it back into the jar, and add five blue marbles to the jar from the supply of extras. Do this two more times. Now, pull a marble from the jar, at random. What's the probability that this last marble is red? What's the probability that we actually drew the same marble all four times?

Accepted Solution

A:

Answer:There is a 57.68% probability that this last marble is red.There is a 20.78% probability that we actually drew the same marble all four times.Step-by-step explanation:Initially, there are 50 marbles, of which:30 are red20 are blueAny time a red marble is drawn:The marble is placed back, and another three red marbles are addedAny time a blue marble is drawnThe marble is placed back, and another five blue marbles are added.The first three marbles can have the following combinations:R - R - RR - R - BR - B - RR - B - BB - R - RB - R - BB - B - RB - B - BNow, for each case, we have to find the probability that the last marble is red. So[tex]P = P_{1} + P_{2} + P_{3} + P_{4} + P_{5} + P_{6} + P_{7} + P_{8}[/tex][tex]P_{1}[/tex] is the probability that we go R - R - R - RThere are 50 marbles, of which 30 are red. So, the probability of the first marble sorted being red is [tex]\frac{30}{50} = \frac{3}{5}[/tex].Now the red marble is returned to the bag, and another 3 red marbles are added.Now there are 53 marbles, of which 33 are red. So, when the first marble sorted is red, the probability that the second is also red is [tex]\frac{33}{53}[/tex]Again, the red marble is returned to the bag, and another 3 red marbles are addedNow there are 56 marbles, of which 36 are red. So, in this sequence, the probability of the third marble sorted being red is [tex]\frac{36}{56}[/tex]Again, the red marble sorted is returned, and another 3 are added.Now there are 59 marbles, of which 39 are red. So, in this sequence, the probability of the fourth marble sorted being red is [tex]\frac{39}{59}[/tex]. So[tex]P_{1} = \frac{3}{5}*\frac{33}{53}*\frac{36}{56}*\frac{39}{59} = \frac{138996}{875560} = 0.1588[/tex][tex]P_{2}[/tex] is the probability that we go R - R - B - R[tex]P_{2} = \frac{3}{5}*\frac{33}{53}*\frac{20}{56}*\frac{36}{61} = \frac{71280}{905240} = 0.0788[/tex][tex]P_{3}[/tex] is the probability that we go R - B - R - R[tex]P_{3} = \frac{3}{5}*\frac{20}{53}*\frac{33}{58}*\frac{36}{61} = \frac{71280}{937570} = 0.076[/tex][tex]P_{4}[/tex] is the probability that we go R - B - B - R[tex]P_{4} = \frac{3}{5}*\frac{20}{53}*\frac{25}{58}*\frac{33}{63} = \frac{49500}{968310} = 0.0511[/tex][tex]P_{5}[/tex] is the probability that we go B - R - R - R[tex]P_{5} = \frac{2}{5}*\frac{30}{55}*\frac{33}{58}*\frac{36}{61} = \frac{71280}{972950} = 0.0733[/tex][tex]P_{6}[/tex] is the probability that we go B - R - B - R[tex]P_{6} = \frac{2}{5}*\frac{30}{55}*\frac{25}{58}*\frac{33}{63} = \frac{49500}{1004850} = 0.0493[/tex][tex]P_{7}[/tex] is the probability that we go B - B - R - R[tex]P_{7} = \frac{2}{5}*\frac{25}{55}*\frac{1}{2}*\frac{33}{63} = \frac{825}{17325} = 0.0476[/tex][tex]P_{8}[/tex] is the probability that we go B - B - B - R[tex]P_{8} = \frac{2}{5}*\frac{25}{55}*\frac{1}{2}*\frac{30}{65} = \frac{750}{17875} = 0.0419[/tex]So, the probability that this last marble is red is:[tex]P = P_{1} + P_{2} + P_{3} + P_{4} + P_{5} + P_{6} + P_{7} + P_{8} = 0.1588 + 0.0788 + 0.076 + 0.0511 + 0.0733 + 0.0493 + 0.0476 + 0.0419 = 0.5768[/tex]There is a 57.68% probability that this last marble is red.What's the probability that we actually drew the same marble all four times?[tex]P = P_{1} + P_{2}[/tex][tex]P_{1}[/tex] is the probability that we go R-R-R-R. It is the same [tex]P_{1}[/tex] from the previous item(the last marble being red). So [tex]P_{1} = 0.1588[/tex][tex]P_{2}[/tex] is the probability that we go B-B-B-B. It is almost the same as [tex]P_{8}[/tex] in the previous exercise. The lone difference is that for the last marble we want it to be blue. There are 65 marbles, 35 of which are blue.[tex]P_{2} = \frac{2}{5}*\frac{25}{55}*\frac{1}{2}*\frac{35}{65} = \frac{875}{17875} = 0.0490[/tex][tex]P = P_{1} + P_{2} = 0.1588 + 0.0490 = 0.2078[/tex]There is a 20.78% probability that we actually drew the same marble all four times