Independent Trials

Toss a coin several times and the number of heads varies each round, yet the probability of each count hides a clear rule. An independent trial repeats the same trial n times with a fixed success probability p, and the chance of exactly r successes out of n is set by the ways to place the successes times the success and failure probabilities. Collect these probabilities by success count and the bell-shaped binomial distribution emerges. Drag the successes-r slider to see which bar lights up and how P(X=r) rises and falls.

Repeat the Same Trial

Successes r2

👀 See It

①An independent trial repeats the same trial independently each time

②The probability of r successes out of n is the height of one bar

③Collecting the probabilities by success count gives the binomial shape

The Independent-Trial Formula

Probability of independent trials

P(X=r) = C(n,r) p^r (1−p)^n-r

r successes out of n — C(n,r) counts the ways to place the successes

A product of three pieces

C(n,r) × p^r × (1−p)^n-r

①placement C(n,r) ②r successes pʳ ③(n−r) failures (1−p)ⁿ⁻ʳ

How It Differs from Independent Events

🔁 The Repeat Count Is Key

①Independent events: joint occurrence P(A∩B)=P(A)P(B)

②Independent trials: the distribution of repeating one trial n times

③If a repeat count n and success count r appear, it is an independent trial

Compute It Directly

Example 1

Find the probability of getting heads exactly twice in 4 coin tosses.

Put n=4, p=1/2, r=2 into the formula.

C(4,2) (1/2)² (1/2)²

Compute.

= 6 × 116 = 38

▸ 3/8

When p=1/2, (1/2)ⁿ is common, so only the coefficient C(n,r) matters.

Example 2

Find the probability of rolling a 6 exactly once in 3 dice rolls.

n=3, p=1/6, r=1.

C(3,1) (1/6)¹ (5/6)²

Compute.

= 3 × 16 × 2536 = 2572

▸ 25/72

Do not drop the power of the failure probability (1−p)=5/6.

Wrap-up

Key result

P(X=r) = C(n,r) p^r (1−p)^n-r

r successes out of n repeats — a product of placement, successes, failures

2020 CSAT Math type, adapted

If a single shot hits with probability 1/3, what is the probability of hitting exactly twice in 5 shots?

①40/243

②80/243

③10/243

④40/81

⑤8/81

▸ ② 80/243

Put n=5, p=1/3, r=2 into the formula.

C(5,2) (1/3)² (2/3)³

Compute.

= 10 × 19 × 827 = 80243

🎯 Exam Points

①P(X=r)=C(n,r)pʳ(1−p)ⁿ⁻ʳ

②C(n,r) places the successes

③the power of the failure prob (1−p) is essential

④distinguish "exactly r" from "at least r" (complement)

⑤distinguish independent events P(A)P(B) from independent trials (repetition)

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Conditional Probability

Probability Distribution

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