SME
Grade 7 Math · Problem type
Expected results in the long run
Use this for “3 of 8 sectors are prizes. In 240 spins, about how many prizes?”
Diagram
P(red) = 3/8
t = 240 (trials)
This is an example. Type your own values to start a new problem.
E
=
f
n
×
t
o
Other sectors
worked out (not typed)
P
Probability
worked out (not typed)
Formulas
o
=
n
−
f
(other
sectors
=
sectors
in
all
−
favorable
sectors)
5 = 8 − 3
P
=
f
÷
n
(probability
=
favorable
sectors
÷
sectors
in
all)
3
8
=
3
÷
8
E
=
f
÷
n
×
t
(expected
count
=
favorable
sectors
÷
sectors
in
all
×
trials)
90 = 3 ÷ 8 × 240
Assumptions
•
The probability says the share of trials in the long run.
•
Expected count = probability × trials; the real count is near it, not exactly it.
•
Past results do not change the next spin: each spin starts fresh.
Step-by-step
WE KNOW
f = 3 (favorable sectors), n = 8 (sectors in all), t = 240 (trials)
FIND
other sectors (o), probability (P), expected count (E)
Step 1 · Find other sectors (o)
Use o = n − f (other sectors = sectors in all − favorable sectors)
The sectors that are not favorable.
o
=
8
−
3
o
=
5
Step 2 · Find probability (P)
Use P = f ÷ n (probability = favorable sectors ÷ sectors in all)
The favorable sectors out of all of them.
P
=
3
8
P
=
3
8
Step 3 · Find expected count (E)
Use E = f ÷ n × t (expected count = favorable sectors ÷ sectors in all × trials)
The
share
of
the
trials:
the
probability,
f
n
,
times
the
trials.
E
=
3
8
×
240
E
=
0.375
×
240
E
=
90
Check
Put the numbers back in. Both sides should match.
5 = 8 − 3 ✓
3
8
=
3
÷
8
✓
90 = 3 ÷ 8 × 240 ✓
Topic check
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