Key Idea: X counts successes. Identify n and p, then use the GDC for the probability asked.
When is it binomial? Fixed attempts · independent · success/failure · constant p. Example: 20 fair die rolls; count sixes → X ~ B(20,1/6).
Mean, variance and SD Mean: np. Variance: np(1 − p). SD: √variance. Example: B(20,0.3) → 6, 4.2, 2.05.
Exactly / at most / fewer than Exactly 7: binompdf(n,p,7). At most 7: binomcdf(n,p,7). Fewer than 7: binomcdf(n,p,6).
At least / more than / between At least 7: 1 − binomcdf(n,p,6). More than 7: 1 − binomcdf(n,p,7). Between 3 and 8: CDF(8) − CDF(2).
Working backwards Expected limit: write np ≤ limit, then solve. Example: 0.9n ≤ 54 → n ≤ 60. Check 61 fails. Probability limit: test integers and check the neighbour.
More than one stage Find p first: two independent checks → 0.8 × 0.9 = 0.72. Expected repeated events: find whole-event probability q; then Nq. Example: 25 batches × q ≈ 7.59 qualifying batches.
IB-style question [6 marks]
An online shop knows that, independently, 20% of all orders are returned. On a particular day it receives 15 orders. Let X be the number of these orders that are returned. (a) Write down the expected number of returned orders. (b) Find the probability that exactly 2 orders are returned. (c) Find the probability that at least 4 orders are returned.
Step by step:
Each order is independent with a constant return probability, so X is binomial with n = 15 and p = 0.2.
(a) The expected number is the binomial mean np.
(b) 'Exactly 2' uses the single-value probability binompdf.
(c) 'At least 4' is the complement of 'at most 3', so subtract the cumulative probability up to 3.
Use binomcdf(15, 0.2, 3), then subtract from 1.
(a) 3 orders. (b) 0.231. (c) 0.352.
Exam checklist State X, n and p. Check whether the boundary is included. Show the event or calculator setup. Keep intermediate values unrounded.