Question 1203301
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This is a good exercise in basic binomial probability.  Calculate the probabilities of 0, 1, 2, or 3 balls in the bucket and verify that the sum of the probabilities is 1.<br>
If Y (yes) represents a ball in the bucket and N (no) represents a ball not in the bucket, then the probability distribution for the game is found using<br>
{{{(0.4Y+0.6N)^3}}}<br>
0 balls....<br>
{{{C(3,0)((0.4)^0)((0.6)^3)=(1)(1)(0.216)=0.216}}}<br>
1 ball....<br>
{{{C(3,1)((0.4)^1)((0.6)^2)=(3)(0.4)(0.36)=0.432}}}<br>
2 balls....<br>
{{{C(3,2)((0.4)^2)((0.6)^1)=(3)(0.16)(0.6)=0.288}}}<br>
3 balls....<br>
{{{C(3,3)((0.4)^3)((0.6)^0)=(1)(0.064)(1)=0.064}}}<br>
The PDT for the game....<pre>
 balls probability
   0    0.216
   1    0.432
   2    0.288
   3    0.064</pre>
CHECK: 0.216 + 0.432 + 0.288 + 0.064 = 1.000<br>