Hello Math central,
My question is. Forty balls numbered 1-10 are placed in a bag and four are drawn at random. What is the probability that the first ball drawn is between 1-10 (inclusive) the second is between 11-20, the third is between 21-30 and the last is between 31-40?
If you help me I would appreciate it.
Thank you,
Sheila
Hi Sheila,
You need to add more information. Do you replace the balls after each draw? If so then the probability of success on a particular draw is 10/40 and you are asking for 4 successes in a row which has probability (1/4)4 = 1/256.
If the balls are not replaced, the answer gets more complicated. Let us know.
Penny.
Hi Penny,
Thank you for responding to me so quickly. The way I wrote the problem, is the way it was given.
Sheila.
Sheila,
If you don't replace the balls after each draw then you can proceed this way.
The probability that the first ball drawn is between 1 and 10 is 10/40 since ten of the forty balls are between 1 and 10.
Now you need one of the balls between 11 and 20 on the second draw. Ten of the remaining balls thirty-nine balls are marked 11 to 20. Thus if you drew between 1 and 10 on the first ball then the probability you draw between 11 and 20 on the second ball is 10/39. Hence the probability you draw between 1 and 10 on the first draw AND between 11 and 20 on the second draw is
10/40
10/39
Now you need one of the balls between 21 and 30 on the third draw. Ten of the remaining balls thirty-eight balls are marked 21 to 30. Thus if you drew between 1 and 10 on the first ball and between 11 and 20 on the second then the probability you draw between 21 and 30 on the third ball is 10/38.
Can you complete the problem?
Penny