HEEELPPP PLEASSSE!!!!!!!!!! I have 6 elements (a, b, c, d, e, f) I close my eyes and randomly pick 5 elements. What is the chance of getting a and b in those 5 elements?
c'mon there must be someone who knows how to do it.. :S
probability isn't my strong point but i think its 2/3 there 6 ways of picking out 5 elements from 6 four of them have no c,d,e or f and include a and b so thats 4/6 or 2/3
how do you calculate that there are four of them have no c,d,e or f
if you randomly picked 5 among 5, the total number of distinct combinations is C(6,5) = 6 if a and b must be among those 5, then choose the two letters C(2,2) = 1 and any three letters among four C(4,3)=4. by multiplication rule, the number of outcomes with this condition is 1(4) = 4 so the probability of the success of this experiment is 4/6 = 2/3
uhm, pick 5 among 6 (not 5).
the probability that this experiment will fail is when you didn't pick a nor b, and the probability of this "failure" is 2/6 = 1/3. THEREFORE the probability that this experiment is a "success" is 1-1/3 = 2/3. Same result. =)
why did you do C(2,2) ?
C(2,2) means you pick the two letters a and b, C(2,1) means pick either a or b, while C(2,0) pick neither a nor b.
another way is to compute \[\frac{\dbinom{4}{3}}{\dbinom{6}{5}}\] you will get the same answer
i am merely showing that the two letters a and b are selected (that's C(2,2)) and three more among 4 (that's C(4,3)) to show in the solution that 5 letters are actually picked.
the denominator, which is just 6, is the number of ways you can choose 5 out of 5, that is the number of elements in the sample space the numerator (which is just 4) is the number of ways you can choose 3 out of the 4 that are not \(a\) and \(b\) this also gives you \(\frac{4}{6}\)
\[\frac{ \left(\begin{matrix}2 \\ 2\end{matrix}\right) \left(\begin{matrix}4 \\ 3\end{matrix}\right) }{ \left(\begin{matrix}6 \\ 5\end{matrix}\right) }=\frac{1 \times 4}{6}=\frac{2}{3}\]
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