SME
Grade 11 Math · Problem type
Binomial: at least k successes
Use this for “A fair coin is tossed 5 times. What is the chance of at least 4 heads?”
Chart
Binomial bars: n = 5 trials, each a success with chance p = 0.5.
E(X) = n × p = 5 × 0.5 = 2.5
σ = √(n × p × (1 − p)) = 1.118
P(X ≥ 4) = P(4) + P(5) = 0.1563 + 0.0313 = 0.1875, the lit bars.
This is an example. Type your own values to start a new problem.
n
Trials
example
p
Chance of success on each trial
example
k
At least this many successes
example
P
Probability of at least k successes, P(X ≥ k)
worked out (not typed)
Formulas
P
=
P
(X
≥
k
)
in
n
trials
at
p
0.1875
=
P
(X
≥
4)
in
5
trials
at
0.5
Assumptions
•
At least k means k, k + 1, … up to n successes: add those bars.
•
Each bar is P(X = j) = C(n, j) × pʲ × (1 − p)ⁿ⁻ʲ, with the same p on every independent trial.
•
When fewer bars lie below k, add those instead and take them from 1: P(X ≥ k) = 1 − P(X < k).
Step-by-step
WE KNOW
n = 5, p = 0.5, k = 4
FIND
probability of at least k successes, P(X ≥ k) (P)
Step 1 · Find probability of at least k successes, P(X ≥ k): P
Use P = P(X ≥ k) in n trials at p
Add
the
bars
from
k
up
to
n
:
each
is
the
orders
of
j
successes,
C(
n
,
j),
times
the
chance
of
one
order.
P
=
5
×
p
4
×
(1 −
p
)
1
+
1
×
p
5
×
(1 −
p
)
0
P
=
5
×
0.5
4
×
(1 − 0.5)
1
+
1
×
0.5
5
×
(1 − 0.5)
0
P
=
0.15625
+
0.03125
P
=
0.1875
Check
Put the numbers back in. Both sides should match.
0.1875
=
5
×
0.5
4
×
(1 − 0.5)
1
+
1
×
0.5
5
×
(1 − 0.5)
0
✓
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