Conditional Probability worksheet with problems on sweets, tiles, and socks, including a tree diagram.

Let’s solve each problem step by step.

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Problem 1:

A bag has 3 red sweets and 4 yellow sweets → total = 7 sweets.

Tom takes one sweet without looking, then passes the bag to Harry, who also picks one without looking.

We are told: Tom chooses a red sweet. So now, what’s left in the bag?

Originally:
Red = 3, Yellow = 4 → Total = 7

After Tom takes 1 red:
Red = 2, Yellow = 4 → Total = 6

Now Harry picks from these 6 sweets.

a. Probability Harry chooses a red sweet:
→ Red left = 2, Total = 6 → P(red) = 2/6 = 1/3

b. Probability Harry chooses a yellow sweet:
→ Yellow = 4, Total = 6 → P(yellow) = 4/6 = 2/3

✔ Check: 1/3 + 2/3 = 1 → makes sense.

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Problem 2:

Word: MATHEMATICS → Let’s count letters:

M A T H E M A T I C S → that’s 11 letters.

List them with counts:

- M: appears 2 times (positions 1 and 6)
- A: appears 2 times (positions 2 and 7)
- T: appears 2 times (positions 3 and 8)
- H: 1 time
- E: 1 time
- I: 1 time
- C: 1 time
- S: 1 time

Total tiles = 11

Tiles are taken out without replacement → so after first pick, only 10 left.

We need probabilities for:

a. Two T’s

First pick must be T → there are 2 T’s out of 11 → P(first T) = 2/11

Then second pick must be T again → now only 1 T left out of 10 → P(second T | first T) = 1/10

So P(two T’s) = (2/11) × (1/10) = 2/110 = 1/55

b. E followed by I

P(first is E) = 1/11 (only one E)

Then P(second is I | first was E) = 1/10 (still one I left, 10 tiles total)

So P(E then I) = (1/11) × (1/10) = 1/110

c. A followed by M

P(first is A) = 2/11 (two A’s)

Then P(second is M | first was A) = 2/10 (still two M’s left? Wait — did we remove an A? Yes. But M’s are untouched → still 2 M’s out of 10 remaining tiles)

Wait — let’s double-check:

Original: M=2, A=2, T=2, others=1 each → total 11

If first tile is A → now A=1 left, M=2 still, total tiles = 10

So P(M on second) = 2/10

Thus P(A then M) = (2/11) × (2/10) = 4/110 = 2/55

d. T and H in any order

This means either:

- First T, then H → OR → First H, then T

Case 1: T then H

P(T first) = 2/11

Then P(H second | T first) = 1/10 (H still there, 10 tiles left)

→ (2/11)(1/10) = 2/110

Case 2: H then T

P(H first) = 1/11

Then P(T second | H first) = 2/10 (T’s still both there)

→ (1/11)(2/10) = 2/110

Total = 2/110 + 2/110 = 4/110 = 2/55

✔ All answers checked.

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Problem 3:

Draw has 20 socks:
Black = 10, White = 6, Brown = 4

One sock chosen randomly without replacement, then another.

Part a: Put into probability tree diagram — already drawn in image, but we’ll use it for calculations.

The tree shows:

First branch: Black (10/20), White (6/20), Brown (4/20)

Then from each, second branches show remaining options.

For example, if first is Black (10/20), then remaining:

Black = 9, White = 6, Brown = 4 → total 19

So second branches from Black:
Black: 9/19, White: 6/19, Brown: 4/19

Similarly, if first is White (6/20), then remaining:

Black = 10, White = 5, Brown = 4 → total 19

Second branches:
Black: 10/19, White: 5/19, Brown: 4/19

If first is Brown (4/20), then remaining:

Black = 10, White = 6, Brown = 3 → total 19

Second branches:
Black: 10/19, White: 6/19, Brown: 3/19

Now calculate:

a. Probability of choosing 2 black socks

= P(first black) × P(second black | first black)
= (10/20) × (9/19) = (1/2) × (9/19) = 9/38

b. A white sock and a brown sock (in any order)

Two cases:

Case 1: White then Brown
P(white first) = 6/20
P(brown second | white first) = 4/19
→ (6/20)(4/19) = 24/380

Case 2: Brown then White
P(brown first) = 4/20
P(white second | brown first) = 6/19
→ (4/20)(6/19) = 24/380

Total = 24/380 + 24/380 = 48/380

Simplify: divide numerator and denominator by 4 → 12/95

c. Two matching socks

That means: both black, or both white, or both brown.

We already have both black: 9/38

Both white:
P(first white) = 6/20
P(second white | first white) = 5/19
→ (6/20)(5/19) = 30/380 = 3/38

Both brown:
P(first brown) = 4/20
P(second brown | first brown) = 3/19
→ (4/20)(3/19) = 12/380 = 3/95

Now add all three:

Convert to common denominator. Let’s use 380.

Both black: 9/38 = (9×10)/(38×10) = 90/380
Both white: 3/38 = 30/380
Both brown: 3/95 = (3×4)/(95×4) = 12/380

Total = 90 + 30 + 12 = 132 / 380

Simplify: divide numerator and denominator by 4 → 33/95

d. Two different socks

Easier way: 1 - P(same socks)

P(same) = 33/95 (from part c)

So P(different) = 1 - 33/95 = (95 - 33)/95 = 62/95

Alternatively, you could add up all mixed pairs, but this is faster and correct.

✔ All answers verified.

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Final Answer:

1.
a. 1/3
b. 2/3

2.
a. 1/55
b. 1/110
c. 2/55
d. 2/55

3.
a. 9/38
b. 12/95
c. 33/95
d. 62/95