SME
Grade 10 Math · Problem type
Two draws without replacement
Use this for “A bag has 5 red and 3 blue marbles. Two are drawn without putting the first back. Find P(red, then red).”
Diagram
Each node’s branches add to 1: 5/8 + 3/8 = 1
P(Red and Red) = P(Red) × P(Red | Red) = 5/8 × 4/7 = 5/14
This is an example. Type your own values to start a new problem.
P(Red,
Red)
=
r
n
×
−
1
−
1
=
5
14
P(R, R)
P(R)
P(Red first)
worked out (not typed)
P(R | R)
P(Red second | Red first)
worked out (not typed)
P(R | B)
P(Red second | Blue first)
worked out (not typed)
Formulas
P(R)
=
r
n
5
8
=
5
8
P(R
|
R)
=
r
− 1
n
− 1
4
7
=
5 − 1
8 − 1
P(R
|
B)
=
r
n
− 1
5
7
=
5
8 − 1
P(R, R) = P(R) × (P(R | R))
5
14
=
5
8
×
4
7
Assumptions
•
The first draw is red with chance r ÷ n.
•
Without replacement one marble is gone: after a red, r − 1 reds are left of n − 1.
•
The second draw depends on the first. Multiply along the path.
Step-by-step
WE KNOW
r = 5, n = 8
FIND
P(Red first) (P(R)), P(Red second | Red first) (P(R | R)), P(Red second | Blue first) (P(R | B)), P(Red, then Red) (P(R, R))
Step 1 · Find P(Red first): P(R)
Use P(R) = r ÷ n
Reds out of all the marbles.
P(R)
=
5
8
P(R)
=
5
8
Step 2 · Find P(Red second | Red first): P(R | R)
Use P(R | R) = (r − 1) ÷ (n − 1)
One red and one marble fewer after a red.
P(R
|
R)
=
5 − 1
8 − 1
P(R
|
R)
=
4
7
P(R
|
R)
=
4
7
Step 3 · Find P(Red second | Blue first): P(R | B)
Use P(R | B) = r ÷ (n − 1)
Every red is left after a blue.
P(R
|
B)
=
5
8 − 1
P(R
|
B)
=
5
7
P(R
|
B)
=
5
7
Step 4 · Find P(Red, then Red): P(R, R)
Use P(R, R) = P(R) × (P(R | R))
Multiply along the path.
P(R,
R)
=
5
8
×
4
7
P(R,
R)
=
5
14
Check
Put the numbers back in. Both sides should match.
5
8
=
5
8
✓
4
7
=
5 − 1
8 − 1
✓
5
7
=
5
8 − 1
✓
5
14
=
5
8
×
4
7
✓
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