Revision as of 22:48, 25 June 2024 by Admin (Created page with "'''Solution: E''' The number of yellows balls drawn has probability distribution: <math display = "block"> P(N=k) = \frac{\binom{5}{k} \binom{7}{3-k}}{\binom{12}{3}}. </math> Calculating the above gives: {| class="table" |+ Probability distribution for the number of yellow balls drawn |- ! k !! P(N=k) |- | 1 || 0.4773 |- | 2 || 0.3182 |- | 3 || 0.0455 |} And therefore the expected value equals: 0.4773 + 0.3182*2 + 0.0455*3 = 1.2502")
Exercise
ABy Admin
Jun 25'24
Answer
Solution: E
The number of yellows balls drawn has probability distribution:
[[math]]
P(N=k) = \frac{\binom{5}{k} \binom{7}{3-k}}{\binom{12}{3}}.
[[/math]]
Calculating the above gives:
| k | P(N=k) |
|---|---|
| 1 | 0.4773 |
| 2 | 0.3182 |
| 3 | 0.0455 |
And therefore the expected value equals:
0.4773 + 0.3182*2 + 0.0455*3 = 1.2502