Amount of money a player to receive

[songoku]

Homework Statement

A bag contains 4 reds, 5 blue and 6 green balls (balls are identical except their colour). Ten balls are drawn, one at a time and being replaced before taking next ball. A player receives $0.5 for every blue ball drawn.
(i) Find the probability he gets more than $2.5
(ii) If 20 players are randomly chosen, find the expected number of players who will receive more than $2.5
(iii) Find the amount a player is most likely to receive

Relevant Equations

Binomial Distribution

I can answer parts (i) and (ii).

For part (iii), this is what I did:
Expected number of blue balls drawn out of 10 trials = n.p = 10 x 1/3 = 10/3

The amount a player is most likely to receive = 10/3 x $0.5 = $1.67

But the answer key is $1.5, not really far from my answer but I wonder is there anything wrong with my working?

Thanks

 

[PeroK]

The amount a player is most likely to receive = 10/3 x $0.5 = $1.67

But the answer key is $1.5, not really far from my answer but I wonder is there anything wrong with my working?

It's impossible to receive $1.67. Winnings are always multiples of $0.50.

 

[songoku]

It's impossible to receive $1.67. Winnings are always multiples of $0.50.

I see, so $1.67 is rounded down to $1.5

If the calculated value is $1.8, then I round it to $2. What about $1.75? Does it mean I can either round it to $1.5 or $2 (both are correct)?

Thanks

 

[PeroK]

I see, so $1.67 is rounded down to $1.5

No. You are looking for the most likely, not the average.

 

[songoku]

No. You are looking for the most likely, not the average.

My idea is to compute P(X = 0) until P(X = 10) and see which has the highest probability. Is there a way which does not involve brute force?

Thanks

 

[PeroK]

My idea is to compute P(X = 0) until P(X = 10) and see which has the highest probability. Is there a way which does not involve brute force?

Thanks

Excel has a binomial distribution function, for example. That should be easy to use.

Note that the binomial distribution has a peak around the mean, so that narrows down the possibilities for the most likely.

 

[songoku]

Thank you very much PeroK

 

[FactChecker]

I see, so $1.67 is rounded down to $1.5

If the calculated value is $1.8, then I round it to $2. What about $1.75? Does it mean I can either round it to $1.5 or $2 (both are correct)?

Thanks

It sounds like you expect the distribution to be symmetric around a mean of $1.75. For a binomial distribution, that is not true except for a probability of 1/2. In general, you should calculate the probabilities at the possible values. For this example, $1.67 is so much closer to $1.5 than to $2 that $1.5 is a good guess. You also seem to be assuming that the average (mean) is directly related to the most likely (mode). For some distributions, they can be very different.

 

[Prof B]

The number that maximizes $f(k)$ is called the mode. If there is a unique local maximum then you can find it as follows: Let $r(k) = f(k)/f(k-1)$. Let $m = \max\{k \,|\, r(k) \geq 1\}$. Note that if $r(m) = 1$ then $m-1$ and $m$ are equally like.

For a binomial distribution $r(k) = \frac{n-k+1}{k} \frac{p}{1-p}$. Please verify that $m = \lfloor np+p\rfloor$