01 · One specified order
For that model, find P(SFSF).
Hint
No combination multiplier is needed for one named order.
Worked solution
9/256.
Understand · explore · practise
Binomial probabilities
Derive the binomial formula by counting success positions, calculate exact-count probabilities and distinguish one ordered path from all qualifying orders.
Before you startBinomial model conditions, combinations and independent probabilities.
01 / Start with one ordered path
An exact success count can occur in several orders.
For X~B(n,p), P(X=r)=C(n,r) pʳ (1−p)ⁿ⁻ʳ.
The formula applies to integer r from 0 through n; other values have probability 0.
The power of p counts successes and the power of 1−p counts failures. The combination factor counts all different choices of success positions. First verify the binomial assumptions.
One order or every order?Explore
Four independent trials each succeed with probability p. X counts successes. Choose p and an exact count r.
For r=0, only FFFF qualifies.
02 / Count choices of success positions
Every order with r successes has the same probability.
Watch: six orders contribute to exactly two
Pause, replay or seek freely. The notes explain the same idea and stay in view.
For four trials with p=1/4, find P(X=2).Worked example
One order SSFF has probability (1/4)²(3/4)²=9/256
The same probability applies to any other arrangement of two successes.
Choose the two success positions in C(4,2)=6 ways
SSFF,SFSF,SFFS,FSSF,FSFS,FFSS.
P(X=2)=6×9/256=27/128
Multiplying by 6 adds the probabilities of six disjoint orders.
02 · Count the positions
How many orders give exactly one success in four trials?
Hint
Choose its one position.
Worked solution
C(4,1)=4: SFFF,FSFF,FFSF,FFFS.
03 / Substitute all three inputs
The failure exponent is n−r.
X~B(6,0.4). Find P(X=2).Worked example
C(6,2)=15
There are 15 ways to choose two success positions.
One order has probability 0.4²×0.6⁴=0.020736
Two successes and four failures.
P(X=2)=15×0.020736=0.31104
Keep precision until the final answer.
03 · All successes
For Y~B(5,0.2), find P(Y=5).
Hint
There is just one order.
Worked solution
0.2⁵=0.00032.
04 · No successes
For Y~B(5,0.2), find P(Y=0).
Hint
All five trials fail.
Worked solution
0.8⁵=0.32768.
05 · Exactly one
For Y~B(5,0.2), find P(Y=1).
Hint
Use five possible success positions.
Worked solution
5×0.2×0.8⁴=0.4096.
04 / Choose the point-probability function
Exactly r is a point probability, not a cumulative total.
For a calculator’s binomial point-probability function, enter the trial number n, success chance p and required count r in the order your calculator labels them. Check whether the output is P(X=r) or P(X≤r). The latter adds several counts.
06 · Spot the wrong output
For B(5,0.2), a display gives 0.73728 when asked for exactly one success. What may have happened?
Hint
Add the probabilities of zero and one.
Worked solution
0.32768+0.4096=0.73728. This is P(Y≤1), so the cumulative function may have been used instead of P(Y=1)=0.4096.
07 · Impossible count
For B(5,0.2), find P(Y=6) and P(Y=1.5).
Hint
Check the support before using a factorial formula.
Worked solution
Both are 0. The count is an integer from 0 through 5.
05 / Find p for the event actually counted
Use the success class, not an unrelated raw outcome.
08 · Biased die
A die is designed so that six is three times as likely as any one other face; the other five faces are equally likely. Find P(six).
Hint
Assign one weight to each ordinary face and three to six.
Worked solution
Total weight=5+3=8, so P(six)=3/8.
09 · Two sixes
Roll that die independently four times. Find P(exactly two sixes).
Hint
Use n=4,p=3/8,r=2.
Worked solution
6×(3/8)²×(5/8)²=675/2048, approximately 0.329590.
10 · Reclassify
For that die, find P(not six) and name a binomial model for the non-six count in four independent rolls.
Hint
Complement the success probability.
Worked solution
P(not six)=5/8. The non-six count is B(4,5/8).
06 / Handle the boundary cases by meaning
Certain outcomes give a degenerate distribution.
11 · p=0
For X~B(7,0), find P(X=0) and P(X=1).
Hint
Success is impossible.
Worked solution
P(X=0)=1 and P(X=1)=0. This avoids any ambiguity from entering 0⁰ on a calculator.
12 · p=1
For X~B(7,1), find P(X=7).
Hint
Every trial succeeds.
Worked solution
1.
If n=0, there are no trials and the success count is 0 with probability 1. For ordinary 0<p<1 and positive n, every integer from 0 through n has positive mass.
07 / Use checks that match the question
Equal success counts do not mean equally likely counts.
13 · Equal count probabilities?
For four fair independent trials, are the five success counts 0,1,2,3,4 equally likely?
Hint
Their numbers of orders are 1,4,6,4,1.
Worked solution
No. The masses are 1/16,4/16,6/16,4/16,1/16. Equal probability of the 16 ordered outcomes does not imply equal probability of the five counts.
14 · Changing p
Could the same formula be used if the success chance changes from trial to trial?
Hint
Equal probabilities for all r-success orders were used in the derivation.
Worked solution
Not as a single B(n,p) formula. Different orders can have different probabilities. Use the correct path probabilities or a suitable alternative model.
08 / One order times the number of orders
Keep the event and its assumptions explicit.
Define the success, check fixed n, common p and independence, then identify r. Calculate the probability of one r-success order and multiply by C(n,r). For an exact count, choose the point-probability calculation and round only at the end.
Section 1 of 8 · Start with one ordered path