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From all the balls in the urn $0.4$ are blue and $0.25$ are red. What is the probability that after picking 3 balls we get a blue and a red ball.

My reasoning is as follows. All the different ways we can pick the balls are $3!$ so we have multiply this by the probability of getting one blue and one red.

$$3!0.4*0.25 = 0.6$$

But I'm not sure because I ran few rows of excel (what I think was similar simulation for this task... not sure) and i got answer around $0.4$.

Please tell me if I'm right and if I'm wrong all the tips are welcome. Thank you :)

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    $\begingroup$ I don't understand the downvoting of this question. $\endgroup$ – drhab Feb 27 '18 at 19:28
  • $\begingroup$ Is the probability of a red ball $.25$ or $.2?$ Also $3!0.4*0.2 = 0.6$ looks like pretty strange arithmetic. $\endgroup$ – saulspatz Feb 27 '18 at 19:28
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    $\begingroup$ Hint: you have three relevant scenarios: two red and a blue, two blue and a red, one red one blue and one other. Do them separately and add. $\endgroup$ – lulu Feb 27 '18 at 19:29
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    $\begingroup$ @drhab I agree. Other than the $.25,.2$ ambiguity the question is reasonably clear and there is a clear, rational, attempt to solve it. Simulating it for verification is an excellent impulse (worth noting that the simulation was done well...if $.25$ is correct I get $.405$ as the true answer). $\endgroup$ – lulu Feb 27 '18 at 19:31
  • $\begingroup$ @lulu Maybe the OP is of Dutch origin (like me). We write 0,4 instead of 0.4. $\endgroup$ – drhab Feb 27 '18 at 19:34
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To elaborate on the solution sketched in the comments:

First of all, I assume that the total number of balls is large enough to ensure that the probabilities do not change significantly as we draw some. If you don't wish to make this assumption, then you need to give us some more information regarding the number of balls.

Note: to avoid the problem you could say that you draw "with replacement". That certainly ensures that the probabilities do not change on subsequent draws.

Second, I will assume that you are after "at least one red and at least one blue." so that, for example, getting two reds and a blue would be a success. This is consistent with your simulation result (again, if you intended something else, you should say so).

Given those assumptions, there are three "good" scenarios: two reds and a blue, two blues and a red, one red one blue and one other. We'll do each in turn:

Scenario I: two reds and a blue. If we fix the order, say $RRB$ then the probability is $.25\times .25\times .4$. As there are three possible orders, the probability here is $3\times .25\times .25\times .4=\boxed {.075}$

Scenario II: two blues and a red. Similar. We get $3\times .25\times .4\times .4=\boxed {.12}$

Scenario III: one red one blue one other. If we fix the order, say $RBX$, the probability is $.25\times.4\times .35$. As there are $3!=6$ possible orders the total probability here is $6\times .25\times.4\times .35=\boxed {.21}$

The final answer (depending on the two assumptions spelled out at the start) is then the sum $$.075+.12+.21=\boxed {.405}$$

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I preassume that the probabilities do not change relevantly by taking out two balls.

Let $B$ denote the event that one of the drawn balls is blue.

Let $R$ denote the event that one of the drawn balls is red.

Then: $$P(B\cap R)=1-P(B^{\complement}\cup R^{\complement})=$$$$1-P(B^{\complement})-P(R^{\complement})+P(B^{\complement}\cap R^{\complement})=1-0.6^3-0.75^3+0.35^3=0.405$$

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  • $\begingroup$ the probability that they are all not blue or red is included in that. $\endgroup$ – Sonny Da Silva-Peters Feb 27 '18 at 19:48
  • $\begingroup$ you would have to subtract $0.35^2$ too. $\endgroup$ – Sonny Da Silva-Peters Feb 27 '18 at 19:49
  • $\begingroup$ @SonnyDaSilva-Peters You are right. I thought all balls were blue or red. I will repair. $\endgroup$ – drhab Feb 27 '18 at 19:49
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as pointed in the comments I forgot to exclude the scenarios of RRR and BBB, RRW, BBW. So the following solution is wrong.

I'll assume that the probability of picking a red ball and a blue ball means the probability of picking at least one red and one blue. So

$$P(\textrm{1 red and 1 blue})= P(\textrm{1 white $\cup$ 0 whites})=P(\textrm{1 white})+P(\textrm{0 white})$$

$$=\frac{{white \choose 1}{blue+red \choose 2}}{white+red+blue \choose3}+\frac{blue+red \choose 3}{white+red+blue \choose 3}$$

Now, suppose there are 20 balls, according to the specifications, 8 are blue, 5 are red and 7 are white, so

$$=\frac{{7 \choose 1}{13 \choose 2}}{20 \choose3}+\frac{13 \choose 3}{20 \choose 3} = 0.7298246 $$

Now, if there are 100 balls, 40 are blue, 25 are red and 35 are white, so

$$=\frac{{35 \choose 1}{65 \choose 2}}{100 \choose3}+\frac{65 \choose 3}{100 \choose 3} = 0.7203463$$

so, you see, the probability is different, depending on the total amount of balls. Are you sure all the information of the question has been given?


side note: If you're not used to $a \choose b$ it means the amount of subsets of size $b$ in a set of size $a$ or "how many ways you can pick $b$ balls in an urn with $a$ balls", ${a \choose b }=\frac{a!}{(a-b)!b!}$

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  • $\begingroup$ This is not correct. Expressions like $\binom {blue+red}2$ include choices of two blue balls, or two red balls. $\endgroup$ – lulu Feb 27 '18 at 20:22
  • $\begingroup$ As I said in the beginning, I was assuming he wanted the probability of at least one Red and one blue, which includes 2 Red and 2 blue. He didn't specify in the question if it was at least 1 Red and 1 blue or Only 1 Red and only 1 blue. $\endgroup$ – Matheus barros castro Feb 27 '18 at 20:49
  • $\begingroup$ Oh, but it's incorrect because I forgot the cases of all Red and all blue! Sorry $\endgroup$ – Matheus barros castro Feb 27 '18 at 20:51
  • $\begingroup$ Exactly. Your method counts $RRW,RRR$, etc. as successes. $\endgroup$ – lulu Feb 27 '18 at 20:52

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