Bernoulli vs Binomial (1 Viewer)

mmmm. · Apr 23, 2021

Suppose there are 100 students, and each of them tosses a coin 10 times, making 1000 tosses. Consider the random variable Y where Y is Bernoulli with ‘success’ being the outcome the coin is on its edge so that p=1/600. Assume the 1000 tosses are independent Bernoulli trials for Y. What is the expected number of coins that will be on their edge over all the student tosses?

I know it says to consider it's a Bernoulli, but would it be a binomial since there is a 3rd option on landing on the edge, and we have the probability for it landing on the edge?

cossine · Apr 23, 2021

"I know it says to consider it's a Bernoulli, but would it be a binomial since there is a 3rd option on landing on the edge, and we have the probability for it landing on the edge?"

No

For Bernoulli and Binomial distribution there can only be two outcomes, success or failure. You cannot have a third outcome.

If you want to deal with multiple outcomes then you would model the probability with a multinominal distribution

Trebla · Apr 23, 2021

The two outcomes you are concerned about are landing on the edge and not landing on the edge. It’s not heads/tails/edge. Effectively the heads/tails is merged into a one event with a probability of 599/600.

CM_Tutor · Apr 24, 2021

Note: This post is meant to explore the problem being presented, rather than to give the answer to the question posed, though that is included as well...

So, as @Trebla says, the event Y is the coin landing on its edge, with

$P(\text{coin lands on edge}) = p = \frac{1}{600}$

and

$P(\text{coin does not land on its edge}) = q = 1 - p = \frac{599}{600}$

Since this is a Bernoulli trial and each is independent, after the 1000 trials conducted by the 100 students

$\begin{align*} P(k \text{ successes}) &= \binom{1000}{k}p^k q^{1000-k} \\ &= \binom{1000}{k} \times \left(\frac{1}{600}\right)^k \times \left(\frac{599}{600}\right)^{1000-k} \\ &= \binom{1000}{k} \times \left(\frac{599}{600}\right)^{1000-k} \times \frac{1}{600^k} \end{align*}$

We can see that each of the probabilities takes the form

$P(k \text{ successes}) = c_k \left(\frac{599}{600}\right)^{1000-k} \qquad \text{where $c_k$ is a constant given by$ \qquad c_k = \binom{1000}{k} \frac{1}{600^k}$

We can then summarise the probabilities for $k$ successes in the 1000 trials in a table:

$\begin{table} \begin{tabular}{c c c c} k & c_k = \frac{1}{600^k}\binom{1000}{k} & P(k \text{ successes}) = c_k\left(\frac{599}{600}\right)^{1000-k} & P \text{ to 6 dec. pl.}) \\ 0 & 1 & c_0\left(\frac{599}{600}\right)^{1000} & 0.188613... \\ 1 & \frac{5}{3} & c_1\left(\frac{599}{600}\right)^{999} & 0.314880... \\ 2 & \frac{111}{80} & c_2\left(\frac{599}{600}\right)^{998} & 0.262575... \\ 3 & \frac{18463}{24000} & c_3\left(\frac{599}{600}\right)^{997} & 0.145826... \\ 4 & \frac{18407611}{57600000} & c_4\left(\frac{599}{600}\right)^{996} & 0.060679... \end{tabular} \end{table}$

... but I can't get the table to behave. So, instead, we can display the probabilities this way:

$\begin{align*} P(0 \text{ successes}) &= \binom{1000}{0}\left(\frac{599}{600}\right)^{1000} &&= \left(\frac{599}{600}\right)^{1000} &&&= 0.188613... \\ \\ P(1 \text{ success}) &= \binom{1000}{1}\left(\frac{599}{600}\right)^{999}\frac{1}{600^1} &&= \frac{5}{3}\left(\frac{599}{600}\right)^{999} &&&= 0.314880... \\ \\ P(2 \text{ successes}) &= \binom{1000}{2}\left(\frac{599}{600}\right)^{998}\frac{1}{600^2} &&= \frac{111}{80}\left(\frac{599}{600}\right)^{998} &&&= 0.262575... \\ \\ P(3 \text{ successes}) &= \binom{1000}{3}\left(\frac{599}{600}\right)^{997}\frac{1}{600^3} &&= \frac{18463}{24000}\left(\frac{599}{600}\right)^{997} &&&= 0.145826... \\ \\ P(4 \text{ successes}) &= \binom{1000}{4}\left(\frac{599}{600}\right)^{996}\frac{1}{600^4} &&= \frac{18407611}{57600000}\left(\frac{599}{600}\right)^{996} &&&= 0.060679... \end{align*}$

$\begin{align*} P(5 \text{ or more successes}) &= 1 - P(\text{0 or 1 or 2 or 3 or 4 successes}) \\ &= 1 - (0.188613... + 0.314880... + 0.262575... + 0.145826... + 0.060679...) \\ &= 0.0274 \quad \text{(3 sig. fig.)} \end{align*}$

We can see, as would be expected for something with a low success probability but a substantial number of trials, that the most likely outcome has a small number of successes, and that probabilities rise from 0 successes to the most likely outcome, and then fall beyond that.

Using the standard formula for an expectation, we can $E(X)$ so long as each outcome and its likelihood is known. Since above gives only the most likely outcomes, this can only give us an estimate:

$\begin{align*} E(X) &= \sum_{i=0}^{1000} i \times P(i \text{ successes}) \\ &= 0 \times P(0 \text{ successes}) + 1 \times P(1 \text{ success}) + 2 \times P(2 \text{ successes}) + ... + 1000 \times P(1000 \text{ successes}) \\ &= 0 \times 0.188613... + 1 \times 0.314880... + 2 \times 0.262575... + 3 \times 0.145826... + 4 \times 0.060679... + ... \\ &\approx 1.52022... \qquad \text{on summing just the first five terms} \end{align*}$

This is an underestimate as there are many missing terms (all small). We could even (crudely) estimate the missing terms by recognising that the probability of 5 or more successes is 0.0274 and that the most likely of these will be 5 successes, so estimating:

$\begin{align*} E(X) &\approx 1.52022... + 5 \times P(5 \text{ or more successes}) \\ &= 1.52022... + 5 \times 0.0274 \\ &= 1.657... \\ &\approx 1.66 \qquad \text{(3 sig. fig.)} \end{align*}$

This is then a good match to the actual expectation for binomials determined from $E(X) = np$ , which is what is sought in this case:

$E(X) = np = 1000 \times \frac{1}{600} = \frac{5}{3} = 1.666666... \approx 1.67 \qquad \text{(3 sig. fig.)}$

Bernoulli vs Binomial (1 Viewer)

mmmm.

Active Member

cossine

Well-Known Member

Trebla

Administrator

CM_Tutor

Moderator

Users Who Are Viewing This Thread (Users: 0, Guests: 1)