Question 1202897
<br>
(i) The probability of getting green on each draw is 4/16 = 1/4, so the probability of not getting green is 3/4.  The probability of not getting green 5 times in a row is<br>
(3/4)^5 = 243/1024<br>
(ii) The probability of getting green on each draw is 1/4; the probability of getting another color is 3/4.  If "Y" (yes) represents drawing a green ball and "N" (no) represents getting a color other than green, then the "probability vector" for each draw is<br>
{{{(1/4)Y+(3/4)N}}}<br>
If the draw is repeated 5 times, the probability vector is<br>
{{{((1/4)Y+(3/4)N)^5}}}<br>
For this question, we want the probability of getting a green ball at least 3 times -- i.e., either 3, 4, or 5 times.<br>
This is binomial probability; the computations are straightforward but tedious, so I'm not going to do them for you.<br>
To get the answer, you need to perform these three calculations and add the results:<br>
3 green: {{{C(5,3)((1/4)^3)((3/4)^2)}}}
4 green: {{{C(5,4)((1/4)^4)((3/4)^1)}}}
5 green: {{{C(5,5)((1/4)^5)((3/4)^0)}}}<br>
Note the answer to (i) can be found using this same kind of binomial probability calculation:<br>
0 green: {{{C(5,0)((1/4)^0)((3/4)^5)}}}<br>
But that calculation simplifies to just (3/4)^5, which is the answer shown above.<br>