Expected product

If we pick $5$ balls from the bag, we're going to end up with one of the six sets below:

$A_1= \left\{2,3,4,5,6 \right\}$

$A_2= \left\{1,3,4,5,6 \right\}$

$A_3= \left\{1,2,4,5,6 \right\}$

$A_4= \left\{1,2,3,5,6 \right\}$

$A_5= \left\{1,2,3,4,6 \right\}$

$A_6= \left\{1,2,3,4,5 \right\}$

where each set has an equal probability of being selected. And that probability is $\frac{1}{6}$ .

The product of the elements of $A_1$ is $720$ .

The product of the elements of $A_2$ is $360$ .

The product of the elements of $A_3$ is $240$ .

The product of the elements of $A_4$ is $180$ .

The product of the elements of $A_5$ is $144$ .

And the product of the elements of $A_6$ is $120$ .

The expected value of a random variable is the weighted average of all possible outcomes.

So the expected value of the product of the balls drawn is equal to

$720(\frac{1}{6})+360(\frac{1}{6})+240(\frac{1}{6})+180(\frac{1}{6})+144(\frac{1}{6})+120(\frac{1}{6})$

$=294$ .

There are $\binom{6}{5} = 6$ possible drawings, each with probability $\frac{1}{6}$ . If ball $n$ is left out, then the product is $\frac{6!}{n}$ . Thus, our expected value is

$\sum_{n=1}^6 \frac{1}{6} \frac{6!}{n} \\= 5! \sum_{n=1}^6 \frac{1}{n} \\ =120(1+(\frac{1}{2}+\frac{1}{3}+\frac{1}{6}) + (\frac{1}{4}+\frac{1}{5})) \\= 120(1+1) + 120\frac{9}{20} \\= 240 + 54 \\= 294.$

By symmetry, it's equally likely whether the ball we leave in the bag is 1, 2, 3, 4, 5, or 6. The products of the possible combinations are:

$1*2*3*4*5 = 120$ $1*2*3*4*6 = 144$ $1*2*3*5*6 = 180$ $1*2*4*5*6 = 240$ $1*3*4*5*6 = 360$ $2*3*4*5*6 = 720$

Since these are equally likely, the expected value is their mean, i.e. $\frac{1}{6}(120+144+180+240+360+720) = \boxed{294}$

We draw five distinct numbers and there are six total. This means we pick all but one number. Since each of the products are distinct, they are equally likely. There are $\dbinom {6}{5} = 6$ ways we can do this, so each product has probability $\dfrac {1}{6}$ .

The possible products are the products of five of the given numbers except one of them. The product of the six numbers is $6! = 720$ . So the possible products are:

$\dfrac {720}{1}, \dfrac {720}{2}, \cdots \dfrac {720}{6}$

Since each has probability $\dfrac {1}{6}$ , the expected value is $\begin{aligned} \displaystyle\sum_{i = 1}^{6} \left( \dfrac {1}{6} \cdot \dfrac {720}{i} \right) &= \dfrac {\displaystyle\sum_{i = 1}^{6} \left( \dfrac {720}{i} \right)}{6} \\&= 120 \cdot \displaystyle\sum_{i=1}^{6} \dfrac {1}{i} \\&= 120 \cdot \left( \dfrac {1}{1} + \dfrac {1}{2} + \cdots + \dfrac {1}{6} \right) \\&= \boxed {294}. \end{aligned}$

There are $6$ possible products, all of which are equally likely to occur. Therefore, the expected value is simply the average of the possible products. To compute the products without much effort, we consider which ball is not chosen. Let the ball not chosen be numbered $n$ ; then the product of the remaining balls is $6!/n.$ So, the expected value is $\dfrac{1}{6}(6!/1+6!/2+6!/3+6!/4+6!/5+6!/6)$ $= 5!/1 + 5!/2 + 5!/3 + 5!/4 + 5!/5$ $= 120 + 60 + 40 + 30 + 24 = \boxed{294}.$

We can see that when 5 balls are chosen, it also means that 1 ball is not chosen, and the probability that any of the 6 numbers is the unchosen ball is undeniably $\frac {1}{6}$ . This makes calculation of the expected value of 5 balls easy, as each of the possible products would simply have a probability of occurring $\frac {1}{6}$ of the time.

Hence, the expected value is:

$\frac {1}{6} (\frac {6!}{1} + \frac {6!}{2} + \frac {6!}{3} + \frac {6!}{4} + \frac {6!}{5} + \frac {6!}{6})$

$=\frac {1}{6} (720 + 360 + 240 + 180 + 144 +120)$

$=\frac {1}{6} \times 1764$

$= 294$

There exists 6 different combinations to choose 5 different-numbered balls within the bag, implying six different scenarios for the product of the selected numbers.

1. Excluding 1 in the product means the product will be 6! , which equals 720 .
2. Excluding 2 in the product means the product will be $\frac{6!}{2}$ , which equals 360 .
3. Excluding 3 in the product means the product will be $\frac{6!}{3}$ , which equals 240 .
4. Excluding 4 in the product means the product will be $\frac{6!}{4}$ , which equals 180 .
5. Excluding 5 in the product means the product will be $\frac{6!}{5}$ , which equals 144 .
6. Excluding 6 in the product means the product will be $\frac{6!}{6}$ , which equals 120 .

Summing those possible products gives us 1764 , however the expected value given the conditions must take into account the six separate scenarios and divide the sum of those products by 6 .

$\frac{1764}{6}$ = 294 , which is the solution.

There are $\binom{6}{1}$ possible products, so our answer is $\frac{5!+\frac{6!}{5}+\frac{6!}{4}+\frac{6!}{3}+\frac{6!}{2}+\frac{6!}{1}}{6}=\boxed{294}$

Choosing five balls from six is essentially the same as leaving one ball of six in the bag, which can be done in one of six ways. Since each of these possibilities is equally probably one can simply sum over six ways of starting with six balls and removing one, then calculate the expectation.

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total=0
for b in range(1,7):
    remaining=range(1,7)
    remaining.remove(b)
    a,b,c,d,e=remaining
    total+=a*b*c*d*e

print total, 1.*total/6

Calculator-less solution

The 6 values are 6!, 6!/2, 6!/3, 6!/4, 6!/5, and 6!/6. Putting these under a common denominator of 60, we get 6!(147)/60(6) as the expected value. Since 6! = 720 and 60(6) = 360, this reduces to 2(147) = 294.

import operator

def prod(factors): return reduce(operator.mul, factors, 1)

ev = 0

for i in range(1,7): ev += prod([w for w in range(1,7) if w != i])

ev = ev/6

To get the expected value we should include all cases so we have here 5 balls out of 6 so 6C5=6, then we have 6 cases, Expected value= [(1X2X3X4X5)+(1X2X3X4X6)+(1X2X3X5X6)+(1X2X4X5X6)+(1X3X4X5X6)+(2X3X4X5X6)]/6=294

The product of all the balls is $6! = 720$ . All possible products of five balls are 720, 360, 240, 180, 144, 120 (determined by dividing 720 by 1, 2, 3, 4, 5, 6). All products are equally likely, so the expected value is the arithmetic mean of the possible products, i.e. 294.

If we are picking 5 balls without replacement, that means one of the 6 balls is left in the bag. There is a 1/6 probability that any of the 6 balls are left out (there is an equal probability for each product). If this is the case (every possibility has an equal chance of happening), the expected value equals the average of the values of the products. The products could be $\frac{720}{1}=720$ , $\frac{720}{2}=360$ , $\frac{720}{3}=240$ , $\frac{720}{4}=180$ , $\frac{720}{5}=144$ , $\frac{720}{6}=120$ . Since $720+360+240+180+144+120=1764$ , the expected value is $\frac{1764}{6}=\boxed{294}$

By symmetry, it's equally likely whether the ball we leave in the bag is 1, 2, 3, 4, 5, or 6. The products of the possible combinations are:

1∗2∗3∗4∗5=120 1∗2∗3∗4∗6=144 1∗2∗3∗5∗6=180 1∗2∗4∗5∗6=240 1∗3∗4∗5∗6=360 2∗3∗4∗5∗6=720 Since these are equally likely, the expected value is their mean, i.e. 1/6(120+144+180+240+360+720)= 294

The answer is 294

A better way to view this problem is to think of it as one ball being left in the bag every trial. Then our expected value becomes $\dfrac{1}{6}(\dfrac{6!}{1} + \dfrac{6!}{2} + \dfrac{6!}{3} + \dfrac{6!}{4} + \dfrac{6!}{5} + \dfrac{6!}{6}) = 294$ our final answer.

There are obviously 6 different products:Each without any of the numbers 1-6. Without 1=6!=720 Without 2=6!/2=360 Without 3=6!/3=240 Without 4=6!/4=180 Without 5=6!/5=144 Without 6=5!=120 Adding all of these up, we then find the mean of the totals to find the expected value. Thus, (720+360+240+180+144+120)/6=294

There are 6 possible products of 5 numbers. Let's name them $P_{1}, P_{2}, P_{3}, P_{4}, P_{5}, P_{6}$ where

$P_{1} = 2 \times 3 \times 4 \times 5 \times 6 = 720$

$P_{2} = 1 \times 3 \times 4 \times 5 \times 6 = 360$

$P_{3} = 1 \times 2 \times 4 \times 5 \times 6 = 240$

$P_{4} = 1 \times 2 \times 3 \times 5 \times 6 = 180$

$P_{5} = 1 \times 2 \times 3 \times 4 \times 6 = 144$

$P_{6} = 1 \times 2 \times 3 \times 4 \times 5 = 120$

Since each product wil only appear once, the probability that the product will appear is $\frac{1}{6}.$

Therefore the expected value of the product is $\frac{P_{1}}{6} + \frac{P_{2}}{6} + \frac{P_{3}}{6} + \frac{P_{4}}{6} + \frac{P_{5}}{6} + \frac{P_{6}}{6} = \frac{720}{6} + \frac{360}{6} + \frac{240}{6} + \frac{180}{6} + \frac{144}{6} + \frac{120}{6} = 120 + 60 + 40 + 30 + 24 + 20 = 294$

The expected value is equivalent to (1+1/2+1/3+1/4+1/5+1/6)(6!)/6 = (147/60)(720)/6 = (147)(12)/6 = 147 * 2 = 294 .

There are 6 balls numbered { 1,2,3,4,5,6}. First, figure out the number of ways to draw 5 balls without replacement from the bag that contains 6 balls. That would be 6 choose 5 or 6 ways. These are the choices: {1,2,3,4,5} {1,2,3,4,6} {1,2,3,5,6} {1,2,4,5,6} {1,3,4,5,6} {2,3,4,5,6}

Therefore, the expected value of the product ( multiply the numbers together) = 1/6 ( 120) + 1/6(144)+ 1/6(180)+1/6(240)+1/6(360)+1/6(720)= 294

There are only $\binom{6}{5}=6$ ways to choose $5$ balls. Therefore, all we have to do is find the product of 6 combinations and divide by 6 to get the expected value.

Case 1 : $2*3*4*5*6=720$

Case 2 : $6*5*4*3*1=360$

Case 3 : $6*5*4*2*1=240$

Case 4 : $6*5*3*2*1=180$

Case 5 : $6*4*3*2*1=144$

Case 6 : $5*4*3*2*1=120$

Now we sum all of them up and divide by six: $\frac{720+360+240+180+144+120}{6}=\boxed{294}$

Solution

• 1 x 2 x 3 x 4 x 5 = 120
• 1 x 2 x 3 x 4 x 6 = 144
• 1 x 2 x 3 x 5 x 6 = 180
• 1 x 2 x 4 x 5 x 6 = 240
• 1 x 3 x 4 x 5 x 6 = 360
• 2 x 3 x 4 x 5 x 6 = 720

Since the expected value is 6 the probability of each value is 1/6

120 x 1/6 + 144 x 1/6 + 180 x 1/6 + 240 x 1/6 + 360 x 1/6 + 720 x 1/6

= 294

The possible products are: 720, 360, 240, 180, 144, 120. The probability of one of those being drawn is the same, 1/6. The expected value is a concept not very much known for most people. If you do not know what it is, search the web for it. the final answer is: $\frac{720+360+240+180+144+120}{6}$ .

we do the product of all options, it will be 120,144,180,240,360,720 the you take all and divides per 6 then your awnser will be 294

Each of the six balls has a chance of $\frac{1}{6}$ of not being drawn, while all five other balls are drawn. Therefore,

$E = \frac{6!}{1 \cdot 6} + \frac{6!}{2 \cdot 6} + \frac{6!}{3 \cdot 6} + \frac{6!}{4 \cdot 6} + \frac{6!}{5 \cdot 6} + \frac{6!}{6 \cdot 6} = \boxed{294}.$

There are only six possible ways to draw five out of six balls: you just don't get one ball and the chance for each ball is the same. You thus have to take the sum of these six products and divide by 6 in order to get $\frac{\text{6 products}}{6} = 4\cdot 5 + 4\cdot 6 + 5\cdot 6 + 2\cdot 4\cdot 5+3\cdot 4\cdot 5 + 4\cdot 5\cdot 6 =294.$

Outcomes are

2 3 4 5 6 
                     Product = 720
                     Probability = 1 / 6
1 3 4 5 6
                     Product = 360
                     Probability =  1 / 6
1 2 4 5 6
                     Product = 240
                     Probability = 1 / 6
1 2 3 5 6
                     Product = 180
                     Probability = 1 / 6
1 2 3 4 6
                     Product = 144
                     Probability = 1 / 6
1 2 3 4 5
                     Product = 120
                     Probability = 1 / 6

So Expexted value is:

$\frac{1}{6} ( 720 + 360 + 240 + 180 + 144 + 120)$

= 294

As there are 6 numbers and 5 can be drawn at a time,the number of ways in which the balls can be drawn are C(6,5)=6. The product of the six numbers = 1 * 2 * 3 * 4 * 5 * 6 = 720. We know that in each of the 6 cases one of the numbers is missing. Thus, the sum of all the cases' products = 720 * (1/1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6) =720 * (147/60). =12 * 147 =1764. The expected value of the products =1764 / 6 =294(Answer).

product of all without 1 is 720 product of all without 2 is 360 The other products obtained are 240, 180, 144, 120 The avearage of this 6 possible values is 294

Case 1: {2,3,4,5,6} Product is 720

Case 2: {1,3,4,5,6} Product is 360

Case 3: {1,2,4,5,6} Product is 240

Case 4: {1,2,3,5,6} Product is 180

Case 5: {1,2,3,4,6} Product is 144

Case 6: {1,2,3,4,5} Product is 120

Each of these cases are equally likely, hence expected value is the average of the products, i.e. 294.

The product of 5 balls drawn is the product of 6 balls divided by the ball still in the bag. The ball in the bag is 1, 2, 3, 4, 5, or 6 with the same probability (= 1/6). Then the expected value of the product of the 5 balls drawn is: \frac{\frac{6!}{1} + \frac{6!}{2} + ... + \frac{6!}{6}}{6} = 294

As quoted from Wikipedia, "The expected value is a weighted average of all possible values."

There are 6 different scenarios that can happen in this case,

2, 3, 4, 5, 6 being drawn

1, 3, 4, 5, 6 being drawn

1, 2, 4, 5, 6 being drawn

1, 2, 3, 5, 6 being drawn

1, 2, 3, 4, 6 being drawn

1, 2, 3, 4, 5 being drawn

The chances of these 6 scenarios happening are equal, $\frac{1}{6}$

Therefore, since the chances of them happening is equal, the weighted average is simply the average of these 6 possible products: 720, 360, 240, 180, 144, 120.

The average is 294, and thus the answer.

There are 6 ways of drawing the five balls..

(720+360+240+180+144+120)/6=294

The expected value of the product is $\frac{1}{6} \sum _{k=1}^6 \frac{6!}{k}$ , as there is an equal chance for any of the six balls to be excluded (resulting in six equally-weighted possibilities). This sum comes out to $\boxed{294}$ .

There are $6$ different ways to draw the balls. We will compute the product of each case and average all the cases.

For 1, 2, 3, 4, 5, the product is $1\times2\times3\times4\times5=120$

For 1, 2, 3, 4, 6, the product is $1\times2\times3\times4\times6=144$

For 1, 2, 3, 5, 6, the product is $1\times2\times3\times5\times6=180$

For 1, 2, 4, 5, 6, the product is $1\times2\times4\times5\times6=240$

For 1, 3, 4, 5, 6, the product is $1\times3\times4\times5\times6=360$

For 2, 3, 4, 5, 6, the product is $2\times3\times4\times5\times6=720$

Taking the average, we get the expected value of the product to be $\dfrac{120+144+180+240+360+720}{6}=\boxed{294}$