Random variables

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1. Discrete random variables and their distributions

Warm-up

Three quick questions from earlier lessons.

Discover

X is the number of heads in two coin tosses. Its values are 0, 1, 2. Which list gives their probabilities?

  1. 14,12,14\tfrac{1}{4}, \tfrac{1}{2}, \tfrac{1}{4}
  2. 13,13,13\tfrac{1}{3}, \tfrac{1}{3}, \tfrac{1}{3}

Key idea

A random variable assigns a number to each outcome.

A discrete random variable XX has a list of values with probabilities P(X=x)P(X = x). Every probability is between 0 and 1, and they sum to 1. Probabilities of events such as X≥2X \ge 2 add the matching values.

Worked example

P(X=0)=0.2P(X = 0) = 0.2, P(X=1)=0.5P(X = 1) = 0.5, XX takes only 0, 1, 2.

P(X=2)=1−0.2−0.5=0.3P(X = 2) = 1 - 0.2 - 0.5 = 0.3
Sum to 1.
P(X≥1)=0.8P(X \ge 1) = 0.8
Everything but 0.

Your turn, guided

Your turn: P(X=0)=0.1P(X = 0) = 0.1, P(X=1)=0.6P(X = 1) = 0.6. Find P(X=2)P(X = 2).

0.30.3
1 − 0.1 − 0.6.

Practice ★

  1. A random variable X takes the values 0, 1 and 2, with P(X = 0) = 29/100 and P(X = 1) = 13/100. Find P(X = 2).
  2. A random variable X takes the values 0, 1 and 2, with P(X = 0) = 14/100 and P(X = 1) = 18/100. Find P(X = 2).
  1. Which list could be the probabilities of a random variable with three values?
    1. 0.1, 0.5 and 4/10
    2. 0.5, 0.5 and 0.1
    3. 0.8, 0.4 and −0.2

Watch out!

Practice ★★

  1. X takes the values 0, 1, 2 with P(X = 0) = 25/100 and P(X = 1) = 15/100. Find P(X ≥ 1).
  2. X takes the values 0, 1, 2 with P(X = 0) = 34/100 and P(X = 1) = 15/100. Find P(X ≥ 1).

Exit ticket

Three questions to check you have got it.

  1. A random variable X takes the values 0, 1 and 2, with P(X = 0) = 29/100 and P(X = 1) = 13/100. Find P(X = 2).
  2. A random variable X takes the values 0, 1 and 2, with P(X = 0) = 14/100 and P(X = 1) = 18/100. Find P(X = 2).
  3. A random variable X takes the values 0, 1 and 2, with P(X = 0) = 18/100 and P(X = 1) = 23/100. Find P(X = 2).

Summary card

Discrete distributions

  • Values with probabilities in [0,1][0, 1] summing to 1.
  • Add probabilities for events.

Watch out: values are not automatically equally likely.

2. Expected value and standard deviation of a random variable

Warm-up

Three quick questions from earlier lessons.

Discover

A raffle ticket wins 10 dollars with probability 0.1 and nothing otherwise. What is the average win per ticket over many tickets?

  1. 1 dollar
  2. 5 dollars

Key idea

The mean and SD of a random variable.

E(X)=μX=∑x P(x)E(X) = \mu_X = \sum x\,P(x), the long-run average. Var(X)=∑(x−μ)2P(x)\text{Var}(X) = \sum (x - \mu)^2 P(x) and SD(X)=Var(X)\text{SD}(X) = \sqrt{\text{Var}(X)}. For XX equal to 1 with probability pp and 0 otherwise: μ=p\mu = p, Var=p(1−p)\text{Var} = p(1 - p).

Worked example

XX: 0, 1, 2 with probabilities 0.3, 0.4, 0.3.

E(X)=0+0.4+0.6=1E(X) = 0 + 0.4 + 0.6 = 1
Σ x P(x).
Var(X)=0.3+0+0.3=0.6\text{Var}(X) = 0.3 + 0 + 0.3 = 0.6
Σ (x − 1)² P(x).
SD(X)=0.6≈0.77\text{SD}(X) = \sqrt{0.6} \approx 0.77
Square root.

Your turn, guided

Your turn: XX = 0, 5 with probabilities 0.6, 0.4. E(X)E(X).

22
5 × 0.4.

Practice ★

  1. X takes the values 0, 1, 2 with probabilities 51/100, 15/100 and 34/100. Find E(X).
  2. X takes the values 0, 1, 2 with probabilities 56/100, 25/100 and 19/100. Find E(X).
  1. A game costs 1 dollar. You win 9 dollars (net) with probability 1/7; otherwise you lose your dollar. Find your expected net gain, in dollars.

Watch out!

Practice ★★

  1. X = 1 with probability 7/10 and X = 0 otherwise. Find the standard deviation of X.
  2. X = 1 with probability 9/10 and X = 0 otherwise. Find the standard deviation of X.

Exit ticket

Three questions to check you have got it.

  1. X takes the values 0, 1, 2 with probabilities 51/100, 15/100 and 34/100. Find E(X).
  2. X takes the values 0, 1, 2 with probabilities 56/100, 25/100 and 19/100. Find E(X).
  3. X takes the values 0, 1, 2 with probabilities 77/100, 10/100 and 13/100. Find E(X).

Summary card

Expected value and SD

E(X)=∑xP(x)E(X) = \sum x P(x); SD=∑(x−μ)2P(x)\text{SD} = \sqrt{\sum (x - \mu)^2 P(x)}.

Watch out: weight by probabilities; include losses.

3. Combining random variables

Warm-up

Three quick questions from earlier lessons.

Discover

Two independent dice each have SD about 1.7. Is the SD of their sum about 3.4?

  1. No, smaller: about 2.4
  2. Yes

Key idea

Means always add; variances add for independent variables.

E(a+bX)=a+bE(X)E(a + bX) = a + bE(X), SD(a+bX)=∣b∣ SD(X)\text{SD}(a + bX) = |b|\,\text{SD}(X). E(X±Y)=E(X)±E(Y)E(X \pm Y) = E(X) \pm E(Y). If XX and YY are independent, Var(X±Y)=Var(X)+Var(Y)\text{Var}(X \pm Y) = \text{Var}(X) + \text{Var}(Y): variances add even for a difference.

Worked example

Morning commute: mean 20, SD 3; evening: mean 25, SD 4; independent.

E(total)=45E(\text{total}) = 45
Means add.
Var=9+16=25\text{Var} = 9 + 16 = 25
Variances add.
SD(total)=5\text{SD}(\text{total}) = 5
Square root.

Your turn, guided

Your turn: SD(X) = 6, SD(Y) = 8, independent. SD(X − Y).

1010
√(36 + 64).

Practice ★

  1. X and Y are independent with SD(X) = 27 and SD(Y) = 36. Find SD(X + Y).
  2. X and Y are independent with SD(X) = 18 and SD(Y) = 24. Find SD(X + Y).
  1. SD(X) = 8. Let Y = 12 + 5X. Find SD(Y).

Watch out!

Practice ★★

  1. X and Y are independent with SD(X) = 24 and SD(Y) = 10. Find SD(X − Y).
  2. X and Y are independent with SD(X) = 72 and SD(Y) = 30. Find SD(X − Y).

Exit ticket

Three questions to check you have got it.

  1. X and Y are independent with SD(X) = 27 and SD(Y) = 36. Find SD(X + Y).
  2. X and Y are independent with SD(X) = 18 and SD(Y) = 24. Find SD(X + Y).
  3. X and Y are independent with SD(X) = 12 and SD(Y) = 16. Find SD(X + Y).

Summary card

Combining random variables

  • Means add (and subtract).
  • Independent: variances add, for sums and differences.

Watch out: never add SDs directly.

4. The binomial distribution

Warm-up

Three quick questions from earlier lessons.

Discover

A coin is tossed 3 times. How many sequences have exactly one head?

  1. 3
  2. 1

Key idea

Count the successes in a fixed number of independent trials.

Binomial setting: nn fixed trials, two outcomes, independent, same probability pp of success. Then P(X=k)=(nk)pk(1−p)n−kP(X = k) = \binom{n}{k}p^k(1 - p)^{n - k}, where (nk)\binom{n}{k} counts the arrangements: (n1)=n\binom{n}{1} = n, (n2)=n(n−1)2\binom{n}{2} = \tfrac{n(n - 1)}{2}.

Worked example

n=5n = 5, p=0.2p = 0.2: P(X=2)P(X = 2).

(52)=10\binom{5}{2} = 10
Arrangements.
0.22×0.83=0.020480.2^2 \times 0.8^3 = 0.02048
One arrangement.
P(X=2)=0.2048P(X = 2) = 0.2048
Multiply.

Your turn, guided

Your turn: n=4n = 4, p=12p = \tfrac{1}{2}. P(X=1)P(X = 1).

14\tfrac{1}{4}
4 × (½)⁴.

Practice ★

  1. A free throw is made with probability 1/6. In 5 independent shots, find P(exactly one is made).
  2. A free throw is made with probability 1/5. In 6 independent shots, find P(exactly one is made).
  1. Which situation fits a binomial model?
    1. The number of heads in 9 tosses of a fair coin
    2. The number of tosses until the first head
    3. The number of aces in 9 cards dealt from one deck

Watch out!

Practice ★★

  1. A seed sprouts with probability 1/4, independently. 6 seeds are planted. Find P(exactly two sprout).
  2. A seed sprouts with probability 1/3, independently. 8 seeds are planted. Find P(exactly two sprout).
  1. Boss problem. A machine part is faulty with probability 1/5, independently. Four parts are tested; X counts the faulty ones.

Exit ticket

Three questions to check you have got it.

  1. A free throw is made with probability 1/6. In 5 independent shots, find P(exactly one is made).
  2. A free throw is made with probability 1/5. In 6 independent shots, find P(exactly one is made).
  3. A free throw is made with probability 1/2. In 7 independent shots, find P(exactly one is made).

Summary card

Binomial probabilities

P(X=k)=(nk)pk(1−p)n−kP(X = k) = \binom{n}{k}p^k(1 - p)^{n - k} in a binomial setting.

Watch out: multiply by the number of arrangements.

5. Mean and standard deviation of a binomial variable

Warm-up

Three quick questions from earlier lessons.

Discover

A fair coin is tossed 100 times. About how many heads do you expect?

  1. 50
  2. 25

Key idea

Shortcuts for the binomial.

If XX is binomial with nn and pp: μX=np\mu_X = np and σX=np(1−p)\sigma_X = \sqrt{np(1 - p)}.

Worked example

n=50n = 50, p=0.3p = 0.3.

μ=15\mu = 15
np.
σ2=10.5\sigma^2 = 10.5
np(1 − p).
σ≈3.24\sigma \approx 3.24
Square root.

Your turn, guided

Your turn: n=200n = 200, p=0.1p = 0.1. The mean.

2020
200 × 0.1.

Practice ★

  1. X is binomial with n = 91 and p = 1/9. Find E(X).
  2. X is binomial with n = 87 and p = 1/8. Find E(X).
  1. X is binomial with n = 70 and p = 1/5. Find SD(X).
  2. X is binomial with n = 71 and p = 1/2. Find SD(X).

Watch out!

Practice ★★

  1. A multiple-choice test has 87 questions with 4 options each. A student guesses every answer. What does E(X) = \frac{87}{4} mean?
    1. Over many such guessed tests, the average number correct is about \frac{87}{4}
    2. The student will get exactly \frac{87}{4} right
    3. \frac{87}{4} is the largest possible score

Exit ticket

Three questions to check you have got it.

  1. X is binomial with n = 91 and p = 1/9. Find E(X).
  2. X is binomial with n = 87 and p = 1/8. Find E(X).
  3. X is binomial with n = 59 and p = 1/5. Find E(X).

Summary card

Binomial mean and SD

μ=np\mu = np, σ=np(1−p)\sigma = \sqrt{np(1 - p)}.

Watch out: the SD is the square root of np(1−p)np(1 - p).

6. The geometric distribution

Warm-up

Three quick questions from earlier lessons.

Discover

You roll a die until the first six. What is the probability that the first six comes on roll 3?

  1. (56)216\left(\tfrac{5}{6}\right)^2 \tfrac{1}{6}
  2. (56)316\left(\tfrac{5}{6}\right)^3 \tfrac{1}{6}

Key idea

Waiting for the first success.

Geometric setting: independent trials with the same pp, repeated until the first success; XX = the number of the trial of the first success. P(X=k)=(1−p)k−1pP(X = k) = (1 - p)^{k - 1}p, μX=1p\mu_X = \dfrac{1}{p}.

Worked example

p=0.25p = 0.25.

P(X=2)=0.75×0.25=0.1875P(X = 2) = 0.75 \times 0.25 = 0.1875
First success on trial 2.
P(X≤2)=1−0.752=0.4375P(X \le 2) = 1 - 0.75^2 = 0.4375
Within 2 trials: complement of 2 failures.
μ=4\mu = 4
1/p.

Your turn, guided

Your turn: p=0.1p = 0.1. The mean number of trials.

1010
1/0.1.

Practice ★

  1. Each call reaches a customer with probability 1/2, independently. Find P(the first success is on call 2).
  2. Each call reaches a customer with probability 1/4, independently. Find P(the first success is on call 2).
  1. A die-like spinner shows 'win' with probability 1/10. On average, how many spins are needed to get the first win?

Watch out!

Practice ★★

  1. Each attempt succeeds with probability 1/7. Find P(the first success comes within the first two attempts).
  2. Each attempt succeeds with probability 1/9. Find P(the first success comes within the first two attempts).

Exit ticket

Three questions to check you have got it.

  1. Each call reaches a customer with probability 1/2, independently. Find P(the first success is on call 2).
  2. Each call reaches a customer with probability 1/4, independently. Find P(the first success is on call 2).
  3. Each call reaches a customer with probability 1/2, independently. Find P(the first success is on call 6).

Summary card

Geometric distribution

P(X=k)=(1−p)k−1pP(X = k) = (1 - p)^{k - 1}p, μ=1/p\mu = 1/p.

Watch out: k−1k - 1 failures before the first success.

Answers: Random variables

1. Discrete random variables and their distributions

A

  1. 2950\frac{29}{50}
  2. 1725\frac{17}{25}

B

  1. 0.1, 0.5 and 4/10

C

  1. 34\frac{3}{4}
  2. 3350\frac{33}{50}

D

  1. 2950\frac{29}{50}
  2. 1725\frac{17}{25}
  3. 59100\frac{59}{100}

2. Expected value and standard deviation of a random variable

A

  1. 83100\frac{83}{100}
  2. 63100\frac{63}{100}

B

  1. 37\frac{3}{7}

C

  1. 2110\frac{\sqrt{21}}{10}
  2. 310\frac{3}{10}

D

  1. 83100\frac{83}{100}
  2. 63100\frac{63}{100}
  3. 925\frac{9}{25}

3. Combining random variables

A

  1. 4545
  2. 3030

B

  1. 4040

C

  1. 2626
  2. 7878

D

  1. 4545
  2. 3030
  3. 2020

4. The binomial distribution

A

  1. 31257776\frac{3125}{7776}
  2. 614415625\frac{6144}{15625}

B

  1. The number of heads in 9 tosses of a fair coin

C

  1. 12154096\frac{1215}{4096}
  2. 17926561\frac{1792}{6561}

D

  1. 45\frac{4}{5}

E

  1. 31257776\frac{3125}{7776}
  2. 614415625\frac{6144}{15625}
  3. 7128\frac{7}{128}

5. Mean and standard deviation of a binomial variable

A

  1. 919\frac{91}{9}
  2. 878\frac{87}{8}

B

  1. 2705\frac{2 \sqrt{70}}{5}
  2. 712\frac{\sqrt{71}}{2}

C

  1. Over many such guessed tests, the average number correct is about \frac{87}{4}

D

  1. 919\frac{91}{9}
  2. 878\frac{87}{8}
  3. 595\frac{59}{5}

6. The geometric distribution

A

  1. 14\frac{1}{4}
  2. 316\frac{3}{16}

B

  1. 1010

C

  1. 1349\frac{13}{49}
  2. 1781\frac{17}{81}

D

  1. 14\frac{1}{4}
  2. 316\frac{3}{16}
  3. 164\frac{1}{64}