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A person can be born on a Monday with probability 1/3 and on any other day with equal probability. What is the probability that 4 randomly chosen persons were born on different days of the week.

6!/4! * (1/9)^4 + 6!/3! * (1/9)^3 * 1/3

denominator = 7^4 

*took two cases where one 4 days are possible but not monday other case is when it might be on a monday and 3 other days the answer seemed not correct

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  • $\begingroup$ It seems unlikely that the denominator is $7^4$ if the days are not equally likely. Do you even need a denominator with all your $\frac19$s in the numerator? My guess is that the answer is $\frac{1800}{9^4}\approx 0.2743$ $\endgroup$ – Henry Feb 19 at 14:56
  • $\begingroup$ @Henry , the denominator is 7^4 is typed by mistake, but how do I get in the numerator 1800, and do I just ignore probability of Monday? $\endgroup$ – Ala Bab Feb 19 at 15:13
  • $\begingroup$ Ala Bab : $(6 \times 5 \times 4 \times 3) + {4 \choose 1} \times 3 \times ( 6 \times 5 \times 4) = 1800$ $\endgroup$ – Henry Feb 19 at 21:50
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You have the correct idea, but the selection of days is incorrect. In the first case, where you have 4 different days, none of which is a Monday, you have

$$6\times5\times4\times3 = {6! \over 2!}$$

possibilities to choose the (ordered) Weekdays ($\neq$ Monday).

In the 2nd case (one Monday in the set), you have

$$6\times5\times4 = {6! \over 3!}$$

ways to select the order of the non-Monday days, and $4$ ways where you put the Monday into that order.

That means your overall probability is

$${6! \over 2!}\left({1 \over 9}\right)^4 + 4\times{6! \over 3!}\left({1 \over 9}\right)^3{1 \over 3} = {360 + 4\times120\times3 \over 9^4}={1800 \over 9^4}={200 \over 9^3} \approx 0.274\ldots$$

I'd like to stress that it is imperative that you use ordered weekdays in the counting, even though the actual condition does not depend on any order. The reason is that those $\left({1 \over 9}\right)^4$ and similar probabilities depend on an order.

To make one specific example, the probability that the first person picked has birthday on Tuesday, the second picked on Wednesday, the third picked on Thursday and the 4th on Friday is $\left({1 \over 9}\right)^4$. The probability that the birthdays are (in order from 1st to 4th picked) (Friday, Thursday, Wednesday and Tuesday) is also $\left({1 \over 9}\right)^4$, and the same is true for the other 22 permutations of those 4 days. Each of those possibilities is a different event, each has probability $\left({1 \over 9}\right)^4$.

If ones makes the error of not considering order, one is lumping all those 24 cases into 1 case. If the used probability is still $\left({1 \over 9}\right)^4$, this is too small by a factor of 24.

EDIT: I see Henry got the same result in a comment, I think that result was added after I looked at the comment (I remember it was there, but shorter).

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Consider the two cases: 1) one is born on Monday and others on other different days but Monday; 2) all are born on different days but Monday.

$$4\cdot \frac13\cdot \frac69\cdot \frac59\cdot \frac49+\frac69\cdot \frac59\cdot \frac49\cdot \frac39=\frac{1800}{9^4}\approx 0.27.$$

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