Binomial Probabilities: Exactly \(k\) Successes in \(n\) Trials

The binomial distribution calculates the probability of obtaining exactly \(k\) successes in a fixed number of independent trials.

Its probability formula combines two ideas:

1. the probability of one particular success–failure pattern;
2. the number of different patterns containing the same number of successes.

If:\[ X=\text{number of successes in }n\text{ trials}, \]

then:\[ \boxed{ P(X=k) = \binom{n}{k} p^k(1-p)^{n-k} } \]

where:

• \(n\) is the number of trials;
• \(k\) is the desired number of successes;
• \(p\) is the probability of success on each trial;
• \(1-p\) is the probability of failure;
• \(\binom{n}{k}\) counts the possible placements of the successes.

Conditions for a Binomial Model

A random variable follows a binomial distribution when four conditions hold.

Fixed number of trials

The number of trials \(n\) is determined in advance.

Two outcome categories

Each trial is classified as:

• success;
• failure.

The original process may have more than two physical outcomes, but they can sometimes be grouped into these two categories.

Constant success probability

Each trial has the same success probability:\[ P(\text{success})=p. \]

Independent trials

The outcome of one trial does not change the success probability of another.

When these conditions hold:\[ \boxed{ X\sim\operatorname{Binomial}(n,p) } \]

Success Is Defined by the Question

In probability, “success” does not necessarily mean a desirable outcome. It simply means the event being counted.

Depending on the question, success could mean:

• winning a small prize;
• receiving a defective product;
• observing a positive result;
• rolling a 6;
• receiving a customer complaint;
• obtaining a head.

The same underlying process can produce different binomial variables depending on how success is defined.

Before applying the formula, define exactly what counts as success.

The Binomial Coefficient

The binomial coefficient is:\[ \boxed{ \binom{n}{k} = \frac{n!}{k!(n-k)!} } \]

It is read as:

“\(n\) choose \(k\).”

It counts the number of ways to choose which \(k\) of the \(n\) trial positions contain successes.

The factorial notation means:\[ n! = n(n-1)(n-2)\cdots2\cdot1. \]

By convention:\[ 0!=1. \]

Example of the Binomial Coefficient

Suppose there are ten trials and exactly two must be successes.

The number of possible success placements is:\[ \binom{10}{2} = \frac{10!}{2!8!}. \]

Expand:\[ 10! = 10\cdot9\cdot8!. \]

Therefore:\[ \binom{10}{2} = \frac{ 10\cdot9\cdot8! }{ 2\cdot1\cdot8! }. \]

Cancel \(8!\):\[ \binom{10}{2} = \frac{10\cdot9}{2} = 45. \]

Thus, there are 45 different patterns containing exactly two successes among ten trials.

Examples include:

SSFFFFFFFF
SFSFFFFFFF
SFFSFFFFFF
...
FFFFFFFFSS

Every pattern has two successes and eight failures.

Probability of One Particular Pattern

Suppose each trial has:\[ P(S)=p \]

and:\[ P(F)=1-p. \]

For a particular pattern containing \(k\) successes and \(n-k\) failures, independence gives:\[ p^k(1-p)^{n-k}. \]

For example, the pattern:

SFSFFFFFFF

has two successes and eight failures. Its probability is:\[ p^2(1-p)^8. \]

The order does not change this probability because every trial has the same success probability.

Why the Binomial Coefficient Is Needed

The expression:\[ p^k(1-p)^{n-k} \]

calculates the probability of only one particular arrangement.

But “exactly \(k\) successes” includes every possible arrangement with \(k\) successes.

There are:\[ \binom{n}{k} \]

such arrangements.

Because distinct complete patterns are mutually exclusive, their probabilities can be added. Since they all have the same probability, this becomes multiplication:\[ P(X=k) = \binom{n}{k} p^k(1-p)^{n-k}. \]

Online Game Example

Suppose an online game is played ten times. On each play:

• big prize: probability 0.10;
• small prize: probability 0.20;
• no prize: probability 0.70.

We want the probability of winning exactly two small prizes.

At first, the process appears not to be binomial because each play has three possible outcomes. However, the question only counts small prizes.

Define:\[ \text{success}=\text{win a small prize} \]

and:\[ \text{failure}=\text{do not win a small prize}. \]

Failure includes:

• winning a big prize;
• winning nothing.

Therefore:\[ p=0.20 \]

and:\[ 1-p = 0.10+0.70 = 0.80. \]

Assuming the ten plays are independent and use the same probabilities:\[ X\sim\operatorname{Binomial}(10,0.20). \]

Calculating Exactly Two Small Prizes

We want:\[ P(X=2). \]

The binomial formula gives:\[ P(X=2) = \binom{10}{2} (0.20)^2 (0.80)^{10-2}. \]

Thus:\[ P(X=2) = \binom{10}{2} (0.20)^2 (0.80)^8. \]

Because:\[ \binom{10}{2}=45, \]

we obtain:\[ P(X=2) = 45(0.20)^2(0.80)^8. \]

Calculate the powers:\[ (0.20)^2=0.04 \]

and:\[ (0.80)^8=0.16777216. \]

Therefore:\[ P(X=2) = 45(0.04)(0.16777216). \]

Thus:\[ P(X=2) \approx0.3019899. \]

As a percentage:\[ \boxed{ P(X=2)\approx30.2\% } \]

Why Big Prizes Are Included in Failure

The question asks only about the number of small prizes.

Therefore, every non-small-prize result belongs to the failure category:\[ P(F) = P(\text{big prize}) + P(\text{nothing}). \]

Because those outcomes are mutually exclusive:\[ P(F) = 0.10+0.70 = 0.80. \]

The binomial model does not distinguish between the two types of failure. It only records whether each play produced a small prize.

If the question separately counted both small and big prizes, a multinomial model would be more appropriate.

A Tree Interpretation

For a small number of trials, the process could be represented with a probability tree:

Trial
├── Small prize: 0.20
└── Not small prize: 0.80

After ten trials, each complete path contains ten branch choices.

Any path with exactly two small-prize branches has probability:\[ (0.20)^2(0.80)^8. \]

There are 45 such paths, giving:\[ 45(0.20)^2(0.80)^8. \]

The binomial formula is a compact alternative to drawing the full tree.

Expected Number of Successes

If:\[ X\sim\operatorname{Binomial}(n,p), \]

then:\[ \boxed{ E[X]=np } \]

For the game:\[ E[X] = 10(0.20) = 2. \]

The expected number of small prizes is two.

This does not mean every ten-play sequence produces exactly two. It means that over many repeated sets of ten plays, the average number approaches two.

It is not a coincidence that \(X=2\) has relatively high probability: two is the expected count in this example.

Variance and Standard Deviation

For a binomial random variable:\[ \boxed{ \operatorname{Var}(X) = np(1-p) } \]

and:\[ \boxed{ \operatorname{SD}(X) = \sqrt{np(1-p)} } \]

For the game:\[ \operatorname{Var}(X) = 10(0.20)(0.80) = 1.6. \]

Therefore:\[ \operatorname{SD}(X) = \sqrt{1.6} \approx1.265. \]

The number of small prizes typically varies around its mean of 2 with a standard deviation of approximately 1.27.

Complete Probability Distribution

For:\[ X\sim\operatorname{Binomial}(10,0.20), \]

the possible values are:\[ X\in\{0,1,2,\ldots,10\}. \]

Each probability is:\[ P(X=k) = \binom{10}{k} (0.20)^k (0.80)^{10-k}. \]

Selected values are:

The probabilities over all possible values sum to one:\[ \sum_{k=0}^{10}P(X=k)=1. \]

Small rounding differences may make a displayed percentage table sum to slightly more or less than 100%.

Exactly, At Most, and At Least

The wording of a binomial question determines which probabilities must be included.

Exactly \(k\)

\[ P(X=k) = \binom{n}{k}p^k(1-p)^{n-k}. \]

At most \(k\)

“At most \(k\)” means:\[ X\leq k. \]

Therefore:\[ \boxed{ P(X\leq k) = \sum_{j=0}^{k} \binom{n}{j} p^j(1-p)^{n-j} } \]

At least \(k\)

“At least \(k\)” means:\[ X\geq k. \]

Therefore:\[ \boxed{ P(X\geq k) = \sum_{j=k}^{n} \binom{n}{j} p^j(1-p)^{n-j} } \]

More than \(k\)

\[ P(X>k) = P(X\geq k+1). \]

Fewer than \(k\)

\[ P(X<k) = P(X\leq k-1). \]

Careful translation of the wording prevents off-by-one errors.

Probability of At Least One Success

The complement of at least one success is zero successes:\[ P(X\geq1) = 1-P(X=0). \]

Since:\[ P(X=0) = (1-p)^n, \]

we obtain:\[ \boxed{ P(X\geq1) = 1-(1-p)^n } \]

For the game:\[ P(X\geq1) = 1-(0.80)^{10}. \]

Therefore:\[ P(X\geq1) = 1-0.107374 = 0.892626. \]

Thus, the probability of at least one small prize is approximately:\[ 89.3\%. \]

Probability of At Most Two Successes

For the game:\[ P(X\leq2) = P(X=0)+P(X=1)+P(X=2). \]

Using the table:\[ P(X\leq2) \approx0.1074+0.2684+0.3020. \]

Therefore:\[ P(X\leq2)\approx0.6778. \]

Thus, there is approximately a 67.8% probability of winning no more than two small prizes.

Probability of More Than Two Successes

Use the complement:\[ P(X>2) = 1-P(X\leq2). \]

Therefore:\[ P(X>2) \approx1-0.6778 = 0.3222. \]

Thus, the probability of winning more than two small prizes is approximately 32.2%.

Multiple Original Outcomes

A process with more than two outcomes can still lead to a binomial model if the outcome of interest can be coded as:\[ I_i= \begin{cases} 1, & \text{if trial }i\text{ is a success}\\ 0, & \text{otherwise}. \end{cases} \]

For the game:\[ I_i= \begin{cases} 1, & \text{small prize}\\ 0, & \text{big prize or no prize}. \end{cases} \]

Then:\[ X=\sum_{i=1}^{10}I_i. \]

Each indicator has:\[ P(I_i=1)=0.20. \]

If the indicators are independent, their sum is binomial.

When the Binomial Model Does Not Apply

Changing success probability

If the game becomes easier or harder after previous plays, \(p\) is not constant.

Dependent trials

If winning one prize changes the probability of later prizes, the trials are not independent.

Sampling without replacement

Drawing items from a small population without replacement changes the remaining composition.

Random number of trials

If play continues until a prize is won, the number of trials is not fixed. A geometric or negative-binomial model may be more suitable.

More than one outcome must be counted separately

If the question asks for both the number of small prizes and the number of big prizes, use a multinomial model.

Different success probabilities

If trial \(i\) has success probability \(p_i\), the count may follow a Poisson-binomial distribution rather than an ordinary binomial distribution.

Common Mistakes

Forgetting the binomial coefficient

\[ p^k(1-p)^{n-k} \]

calculates one arrangement, not all arrangements.

Using the wrong failure probability

Failure means every outcome not classified as success:\[ P(F)=1-p. \]

Using the probability of any prize

If success is a small prize, then:\[ p=0.20, \]

not:\[ 0.10+0.20=0.30. \]

Using the formula when trials are dependent

The factorization into powers relies on independence.

Confusing exactly with at least

\[ P(X=2) \]

is not the same as:\[ P(X\geq2). \]

Treating expected value as the guaranteed outcome

\[ E[X]=2 \]

does not mean exactly two prizes will occur.

Assuming three physical outcomes prevent a binomial model

The process can be binomial if the question reduces outcomes to success versus not success.

A Reliable Binomial Process

1. Define success.
2. Verify that the number of trials is fixed.
3. Verify that trials are independent.
4. Verify that the success probability is constant.
5. identify \(n\), \(k\), and \(p\).
6. calculate the number of arrangements:\[ \binom{n}{k}. \]
7. calculate one arrangement’s probability:\[ p^k(1-p)^{n-k}. \]
8. Multiply the two quantities.
9. Check whether the wording says exactly, at most, or at least.
10. Interpret the result in context.

Key Takeaway

A binomial probability combines the number of ways to place \(k\) successes among \(n\) trials with the probability of each success–failure pattern:\[ P(X=k) = \binom{n}{k} p^k(1-p)^{n-k}. \]

In ten independent game plays where the probability of a small prize is 0.20:\[ P(X=2) = \binom{10}{2} (0.20)^2 (0.80)^8 \approx30.2\%. \]

Although each play has three original outcomes, the question is binomial because success is defined as a small prize and every other outcome is grouped as failure.