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Grade 11-12 (age 16-18)

Independent Trials

Independent Trials

Repeat the Same Trial
2
👀 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) pr (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) × pr × (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.
1
Put n=4, p=1/2, r=2 into the formula.
C(4,2) (1/2)2 (1/2)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.
1
n=3, p=1/6, r=1.
C(3,1) (1/6)1 (5/6)2
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) pr (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
1
Put n=5, p=1/3, r=2 into the formula.
C(5,2) (1/3)2 (2/3)3
2
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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