A six sided unbiased die with four green faces and two red faces is rolled seven times. Which of the following combinations is the likely outcome of the experiment?

(A) Three green faces and four red faces.

(B) Four green faces and three red faces.

(C) Five green faces and two red faces.

(D) Six green faces and one red face.

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My Try:

As probability of occurrence of green face $= 4/6 = 2/3$, and probability of occurrence of red face $= 2/6 = 1/3$.

If the die is rolled $6$ times, then expectation of green and red faces are $4$ and $2$ respectively. But die is rolled $7$ times, so I prefer $5$ green and $2$ red faces as green faces has more probability than red faces.

Could you please explain it?

Thanks in advance.

  • 1
    $\begingroup$ I suspect the question intends you to calculate the probability of the event $4G,3R$ in any order for $C$ and similarly for $D$. What you’re interpreting seems considerably different. $\endgroup$ – Macavity Feb 5 '18 at 7:12
  • $\begingroup$ @Macavity, please check my edit. $\endgroup$ – 1 0 Feb 5 '18 at 7:32
  • 1
    $\begingroup$ If that is the question, as a multiple choice, then the question is wrong. All of the options are possible, some more probable than others, but none are guaranteed to occur / not occur. Do you have a link to the question paper or is this from memory? $\endgroup$ – Macavity Feb 5 '18 at 7:42
  • 2
    $\begingroup$ Yes, this question is wrong, Do you mean which is most likely? $\endgroup$ – Thomas Andrews Feb 5 '18 at 7:53
  • $\begingroup$ And (a) and (d) are now the same. $\endgroup$ – Thomas Andrews Feb 5 '18 at 7:54

The probability of getting 3 red and 4 green. $$\Pr(R=3, G=4)=\binom{7}{3}\left(\frac13\right)^3\left(\frac23\right)^4$$ The probability of getting 2 red and 5 green. $$\Pr(R=2, G=5)=\binom{7}{2}\left(\frac13\right)^2\left(\frac23\right)^5$$

First we compute number of ways in which you can have 3 red and 4 green outcomes in 7 outcomes. This is given by $$\binom73$$now we multiply this with probability that a single outcome may have 3 red and 4 green in a fixed order. $$\left(\frac13\right)^3\left(\frac23\right)^4$$ And we multiply these two. Similarly you can solve for the second one.

For option (C) the probability is $\frac{560}{3^7}$ and for (D) is $\frac{672}{3^7}$ so answer is D

  • $\begingroup$ Thank you for applying Binomial' theorem, but what is answer? $\endgroup$ – 1 0 Feb 5 '18 at 7:16
  • $\begingroup$ $\Pr(R=3, G=4)=\binom{7}{3}\left(\frac13\right)^3\left(\frac23\right)^4$ $\endgroup$ – Sonal_sqrt Feb 5 '18 at 7:18
  • $\begingroup$ Do you mean $7C4 > 7C5?$ $\endgroup$ – 1 0 Feb 5 '18 at 7:23
  • $\begingroup$ Please check my edit. $\endgroup$ – 1 0 Feb 5 '18 at 7:32

@Piyush's answer is correct, but here's a comment that you can settle the issue without a calculator. We want to see, for example, that the probability of event D is greater than the probability of event C. So we want to see, as painlessly as possible, why $$\binom 75\big(\frac23\big)^5\big(\frac13\big)^2 > \binom 74\big(\frac23\big)^4\big(\frac13\big)^3.$$ Note that $\binom 75 = 21$ and $\binom 74 = 35$, and $\frac{21}{35}<\frac12$. On the other hand, to go from $\big(\frac23\big)^5\big(\frac13\big)^2$ to $\big(\frac23\big)^4\big(\frac13\big)^3$, we multiply by a factor of $\frac13\cdot\frac32 = \frac12$. Thus, going from the left-hand side to the right hand side we multiply by a factor $<1$. Thus, the left-hand side is greater.

  • $\begingroup$ Thanks for nice explanation. $\endgroup$ – 1 0 Feb 6 '18 at 5:31

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