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| rev | 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") | Jun 25'24 at 23:48 | +444 |