JC2 H2 Mathematics: Binomial Distribution

Count flavoured chocolates in one packet

A large production run contains 20% mint chocolates. A packet contains 20 randomly selected chocolates. Let X be its number of mint chocolates. State the model assumptions, then find the probability of exactly four mint chocolates.

Assume that whether each chocolate is mint is independent of the others, and that the probability of being mint is constant at 0.2 for every selection. A small sample from a very large mixed run makes these assumptions plausible; a shared batch fault could invalidate independence.

Selecting chocolates until the first mint one appears is a different experiment: the number of selections is not fixed, so that waiting count is not binomial. Sampling a substantial fraction of a finite bag without replacement also changes the success probability.

At least three; more than four but not more than eight

For the same packet, find $P(X\geq3)$ and $P(4\lt X\leq8)$. A cumulative binomial calculation returns $P(X\leq k)$.

The second event contains 5, 6, 7 and 8. Subtracting $P(X\leq3)$ would wrongly include four mint chocolates.

Fewer mint than plain, given at least three mint

Find the probability that a packet has fewer mint than plain chocolates, given that it has at least three mint chocolates.

Together with the given condition, the numerator event is $3\leq X\leq9$.

Here q is the requested conditional probability. The denominator is the probability of the given event alone. The strict inequality excludes the 10–10 tie.

Recover n and p from the mean and variance

A binomial variable has mean 6 and variance 4.2. Find n and p, then the probability of no successes.

Check that n is a positive integer and $0\leq p\leq1$. An impossible parameter value means the proposed binomial model or the input equations must be reconsidered.

An unknown probability inside a mixed pack

A pack contains two toys with independent fault probability 0.02 and three keychains with independent fault probability p. All five fault events are independent. Given at most one faulty item in a pack, the probability that its only fault is a toy is 0.035. Find p.

Let A count faulty toys and K count faulty keychains. The denominator contains only the mutually exclusive cases $(A,K)=(0,0),(1,0),(0,1)$:

Write the numerator N and denominator D separately to keep the conditional equation clear:

The conditioning event has positive probability, so $p\ne1$. Divide by $(1-p)^2$, then solve:

A single binomial model for all five items would be invalid because toys and keychains do not have a common fault probability.

Find the least sample size

Each chocolate is independently mint with probability 0.2. Find the least n such that the probability of fewer than three mint chocolates in n selections is at most 0.3.

Use a table with n restricted to integers. The boundary values are:

Thus the least n is 18. As n increases, adding an independent trial cannot lower the success count, so $P(X\leq2)$ decreases. The preceding integer verifies minimality.

Ten packets: every packet versus the total

A box contains ten independent packets of 20 chocolates. Find the probability every packet has at least three mint chocolates, then the probability the box contains at least 30 mint chocolates.

For one packet, $q=P(X\geq3)=0.793915281\ldots$. Every packet qualifies only when all ten independent packet events occur:

For the second event count all 200 individual chocolates. Their success probability is still 0.2:

Every packet having three mint chocolates guarantees at least 30 in total. The converse fails: packets may compensate for each other. This explains the larger second probability.

If asked instead for exactly eight qualifying packets, define $Y\sim B(10,q)$:

The 10 trials are now packets, and their success probability is q, not 0.2.

The eighth selection is the sixth blue

Balls are selected with replacement, independently, with blue probability 0.6. Find the probability that selection 8 is the sixth blue selection.

Exactly five of the first seven must be blue, and selection 8 must itself be blue. Its position is fixed, so the combination applies only to the first seven:

If the question additionally says the sixth and last blue in ten selections, selections 9 and 10 must both be non-blue: multiply this result by $0.4^2$.

An inspection may require a second sample

Each independent sample contains 10 items with defect probability 0.1 per item. Accept immediately if the first sample has at most one defect; reject if it has at least three. With exactly two, take a second independent sample and accept only if it has at most one defect. Find the acceptance probability, then the chance a second sample was needed given acceptance.

Here $q=P(\text{accept})$ and $r=P(\text{second}\mid\text{accept})$. The branch that triggers retesting is not itself acceptance. It must be followed by a successful second inspection.

Find a range of p for a unique mode

For $X\sim B(5,p)$, find the range of p for which 2 is the unique most probable count. Compare its two neighbours, using $0\lt p\lt 1$.

Hence $\boxed{1/3\lt p\lt 1/2}$. Binomial probabilities increase up to a mode and then decrease; the adjacent ratios $\frac{(5-k)p}{(k+1)(1-p)}$ decrease with k, so these two comparisons establish the global maximum.

At $p=1/3$, counts 1 and 2 tie. At $p=1/2$, counts 2 and 3 tie. If “2 is a mode” allows a tie, include both endpoints; “unique” excludes them.