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Let the first 5 balls in the box be black and the 3 remaining balls red. In turn take out 2 balls. Chance to pull out a (black-red) combination. $P(AB)=P(A|B)P(B)=5/8*3/7=15/56$. (A-pulled the red ball, B-pulled the black ball. But AB intersection of events. event A and B consists of outcomes $A=(6,7,8), B=(1,2,3,4,5)$, hence the intersection is 0 and probability is 0. where am i wrong?

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You just need proper definitions. Let $X_1,X_2$ be your outcomes at the first and second draws, and let $A=\{6,7,8\},B=\{1,2,3,4,5\}$. For example, $X_1=1$, $X_2=6$, would correspond to a black ball in the first draw, and red ball in the second. A black-red combination, without an order, would be $\{(X_1\in A\cap X_2\in B)\cup (X_1\in B\cap X_2\in A)\}$, i.e. black/red or red/black. So, the probability of having black/red combination in the end will be: $$P(X_1\in A,X_2\in B)+P(X_1\in B,X_2\in A)=\frac{3\times5}{8\times7}+\frac{5\times3}{8\times7}=\frac{15}{28}$$

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  • $\begingroup$ well, I'll put it another way.$P(A|B)=card(A*B)/card(B)$. intuitively, it is obvious that the probability is 3/7.but I don't understand how to find $card (AB), card(B)$ technically if set A consists of elementary outcomes $(6,7,8)$ and set B consists of elementary outcomes $(1,2,3,4,5)$.Can you explain this point to me? I apologize for my stupidity $\endgroup$ Commented Oct 27, 2019 at 11:52
  • $\begingroup$ No problem. In your explanation, $A$ is the set of red balls, and $B$ is the set of blacks. However, you don't have a definition for $i$-th draw, I used $X_i$ for it. In simple problems, $P(A),P(B)$ may refer to probabilities of drawing red/black balls, however when we get into multiple draws, you need to introduce more variables to be precise. Otherwise, $P(A),P(B),P(A|B)$ doesn't make sense, e.g. does $P(A|B)$ refer to first black, then red or first red, then black? To cope with it, you need $X_i$. $\text{card}(A)$ doesn't make sense either. $\endgroup$
    – gunes
    Commented Oct 27, 2019 at 12:33
  • $\begingroup$ I have 2 more questions. 1) can I write like this?: event $B$-after the first draw, there are 4 black and 3 red balls in the box. Event $A$-after draw 2, a red ball was taken. Then the set B consists of 7 elements. the intersection of $A$ and $B$ consists of 3 elements. Hence, $P (A |B)=3/7$. 2) could you write {(X1∈A∩X2∈B)=... as in this example (A=(4,6,8,10), B=(4,6), common elements (4,6) therefore A∩B=(4,6)) $\endgroup$ Commented Oct 27, 2019 at 13:52
  • $\begingroup$ 1. if event $B$ is "after the first draw, there are 4 black and 3 red balls in the box", it is actually drawing a "black ball in the first draw". Then, the set $B$ doesn't contain $7$ elements; instead it contains $5$: draw ball 1, ball 2, ..., ball 5. The sample space had $8$ elements, which is why $P(B)=5/8$. 2.) I don't understand your $A$ and $B$ sets here. It's better for you to think as $X_1=R,X_1=B$ etc. or let $A$ and $B$ your first and second draws instead of red/black. $\endgroup$
    – gunes
    Commented Oct 27, 2019 at 19:29

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