IB DP AA Math – Topic: SL 4.8- Binomial distribution, its mean and variance HL Paper 2

IB DP Maths AA : HL Paper -2 : All Topics

Question

In Happyland, the weather on any given day is independent of the weather on any other day. On any day in May, the probability of rain is 0.2. May has 31 days.

Find the probability that
(a) it rains on exactly 10 days in May;
(b) it rains on at least 10 days in May;
(c) the first day that it rains in May is on the 10th day.

▶️Answer/Explanation

Detailed Solution

The number of rainy days in May using a binomial distribution:

X∼Bin(n=31,p=0.2)X \sim \text{Bin}(n=31, p=0.2)

where:

  • XX

    is the number of rainy days,

  • n=31n = 31

    (total days in May),

  • p=0.2p = 0.2

    (probability of rain on a given day).

The binomial probability mass function (PMF) is:

P(X=k)=(nk)pk(1−p)n−kP(X = k) = \binom{n}{k} p^k (1 – p)^{n-k}

(a) Probability that it rains on exactly 10 days

P(X=10)=(3110)(0.2)10(0.8)21P(X = 10) = \binom{31}{10} (0.2)^{10} (0.8)^{21}

Using a calculator:

(3110)=31!10!(31−10)!=30045015 \binom{31}{10} = \frac{31!}{10!(31-10)!} = 30045015 (0.2)10≈1.024×10−7,(0.8)21≈0.1094 (0.2)^{10} \approx 1.024 \times 10^{-7}, \quad (0.8)^{21} \approx 0.1094 P(X=10)=30045015×(1.024×10−7)×0.1094 P(X = 10) = 30045015 \times (1.024 \times 10^{-7}) \times 0.1094 P(X=10)≈0.190P(X = 10) \approx 0.190

So, the probability that it rains exactly 10 days is

0.190\mathbf{0.190}

.

(b) Probability that it rains on at least 10 days

We need to find

P(X≥10)P(X \geq 10)

, which is:

P(X≥10)=1−P(X≤9)P(X \geq 10) = 1 – P(X \leq 9)

Using binomial cumulative probability tables or a calculator, we get:

P(X≤9)≈0.589P(X \leq 9) \approx 0.589 P(X≥10)=1−0.589=0.411P(X \geq 10) = 1 – 0.589 = 0.411

So, the probability that it rains on at least 10 days is

0.411\mathbf{0.411}

.

(c) Probability that the first rainy day is on the 10th day

The first rainy day on the 10th day means that the first 9 days are dry, and the 10th day is rainy.

This follows a geometric distribution:

P(first rain on day 10)=(1−p)9⋅p P(\text{first rain on day 10}) = (1 – p)^9 \cdot p P=(0.8)9×0.2P = (0.8)^9 \times 0.2 ≈0.1342×0.2 \approx 0.1342 \times 0.2 ≈0.0268\approx 0.0268

So, the probability that the first rainy day is on the 10th day is

0.0268\mathbf{0.0268}

.

……………………….Markscheme………………………

Solution: –

6. let X be the number of days of rain in May

(a) recognition of binomial distribution
$X \sim B(31, 0.2)$ or $C_{10}^{31} (0.2)^{10} (0.8)^{21}$ or $X \sim B(n, p)$ or “C, p^{r}(1-p)^{n-r}”
$P(X=10) = 0.0418894…$
= 0.0419

(b) recognition of need to find $P(X \geq 10) = 1 – P(X \leq 9)$
= 0.0745998… (=1-0.925400…)
= 0.0746

(c) recognition of 9 days with no rain followed by a day of rain
$0.8^9 \times 0.2 = 0.0268435…$
= 0.0268

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