Probability question

**noob mathematician** · Mar 24th 2009, 08:42 AM

I have 4 assignments to pass to three different people to do for me. What is the probability that each person will receive at least one assignment?

Total combination: ${4+3-1\choose 3-1}$ = ${6\choose 2}$ = 15

P(each person receive at least one assignment) = $3\over 15$ = $1\over 5$

My answer is 0.2
but it's wrong

It's a MCQ question with options: (A) $8\over 9$ , (B) $64\over 81$ , (C) $4\over 9$ , (D) $16\over 81$ , (E) $5\over 9$

**Plato** · Mar 24th 2009, 09:25 AM

Originally Posted by noob mathematician

I have 4 assignments to pass to three different people to do for me. What is the probability that each person will receive at least one assignment? It's a MCQ question with options: (A) $8\over 9$ , (B) $64\over 81$ , (C) $\color{blue}4\over 9$ , (D) $16\over 81$ , (E) $5\over 9$

Consider that the 4 assignments are all distinct.
There are 36 surjective (onto) functions from a set of 4 to a set of 3.
There are $3^4$ functions from a set of 4 to a set of 3.

**noob mathematician** · Mar 24th 2009, 06:59 PM

Thanks for the reply.

Ya if the assignments are distinct, there will be a total of $3^4$ possibilities.

But how do you find the 36?

My interpretation will be to permute 4 assignments among the 3 people (since each of them will receive one assignment), which is 4. 3. 2. 3 = 72, since the last assignment could be allocated to any one among the three. Therefore answer is A?

**Plato** · Mar 25th 2009, 02:45 AM

Originally Posted by noob mathematician

But how do you find the 36?

The number of mappings from {1,2,3,4} to {a,b,c} which are onto is 36.
Onto means that each of a, b, & c gets at least one.
Look up the ways to count surjections.

**noob mathematician** · Mar 25th 2009, 05:15 AM

oic. Thanks a lot.

The formula S(n,k)= $\sum_{i=1}^{n} {k\choose i}<br /> (k-i)^n (-1)^i$

Ok!

**Plato** · Mar 25th 2009, 06:44 AM

Originally Posted by noob mathematician

The formula S(n,k)= $\sum_{i=1}^{n} {k\choose i}(k-i)^n (-1)^i$

That is close. But not exact.
The number of surjections from a set of N elements to a set of K elements is:
$S(N,K) = \sum\limits_{j = 0}^K {\left( { - 1} \right)^j { K \choose j} \left( {K - j} \right)^N }$ .

**noob mathematician** · Mar 25th 2009, 09:07 PM

Oh ya.. haha some typo error there. Thanks for pointing out.

Thread: Probability question

LinkBack

Thread Tools

Display

Probability question

Similar Math Help Forum Discussions

The desired exam question paper (probability question)

Probability question involving (A given B type question )

Probability Question

Probability Question

A probability question and an Expectations question

Search Tags