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Grade 11-12 (age 16-18)

high school Independent Trials

Toss a coin several times and the number of heads changes every time, yet the probability of each count follows a clear rule. Repeating the same trial n times, each with success probability p and without any trial affecting the others, is called independent trials; the probability of exactly r successes is the number of ways to place the successes, C(n, r), times the success probability pʳ and the failure probability (1 − p)ⁿ⁻ʳ. Lay these probabilities out by number of successes and you get the humped shape of the binomial distribution. Move the slider for the number of successes r and an example arrangement of the successes changes along with the matching bar, while the product of the three pieces is written out under the figure.

Repeating the Same Trial

2
👀 Reading the figure
①Independent trials repeat the same trial so that no result affects the next
②The six circles at the top are the 6 trials and the filled ones are successes; r successes can be placed in C(6, r) ways
③One arrangement has probability (1/3)ʳ(2/3)⁶⁻ʳ, so a bar's height is the number of arrangements times this value
④Putting the bars for each count together gives the binomial shape, and in this figure r = 2 is the tallest

The Formula for Independent Trials

Probability in independent trials
P(X=r) = C(n,r) pr (1−p)n-r
The probability of r successes in n trials. C(n, r) counts the ways to choose where the successes go; the next chapter writes the same formula with ₙCₖ
Product of three pieces
C(n,r) × pr × (1−p)n-r
①choose the places C(n, r) ②r successes pʳ ③n − r failures (1 − p)ⁿ⁻ʳ

How This Differs from Independent Events

🔁 The number of repetitions is the key
①Independent events are two events whose joint probability is P(A ∩ B) = P(A)P(B)
②Independent trials deal with the probability of the number of successes when the same trial is repeated n times
③When a problem gives a number of repetitions n and a number of successes r, use the probability for independent trials

Worked Examples

Example 1
A coin is tossed 4 times. Find the probability of getting heads exactly 2 times.
1
Put n = 4, p = 1/2, r = 2 into the formula.
C(4,2) (1/2)2 (1/2)2
2
Multiply the three pieces.
= 6 × 116 = 38
▸ 3/8
When p = 1/2, pʳ(1 − p)ⁿ⁻ʳ is always (1/2)ⁿ, so only C(n, r) needs comparing.
Example 2
A die is rolled 3 times. Find the probability that a 6 comes up exactly 1 time.
1
Here n = 3, p = 1/6, r = 1.
C(3,1) (1/6)1 (5/6)2
2
Multiply the three pieces.
= 3 × 16 × 2536 = 2572
▸ 25/72
Forget the square of 5/6, the probability of not rolling a 6, and you get the wrong answer 1/2.

Summary

Key result
P(X=r) = C(n,r) pr (1−p)n-r
The probability of r successes in n trials is the product of three pieces: places, successes and 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
Multiply the three pieces.
= 10 × 19 × 827 = 80243
🎯 Exam Points
①P(X = r) = C(n, r) pʳ(1 − p)ⁿ⁻ʳ
②C(n, r) is the number of ways to choose where the successes go
③Do not leave out the power of the failure probability (1 − p)
④Tell "exactly r times" apart from "at least r times"; "at least" is often quicker through the complement
⑤Distinguish independent events, which multiply P(A)P(B), from independent trials, which deal with repetition
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Conditional Probability
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Probability Distribution

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