Probability of Combined Events (Solutions)

Probability of Combined Events — Solutions

Full worked solutions, with method and final answer for every part.

Question 1

(a)There are 36 equally likely outcomes. The pairs summing to 7 are (1,6),(2,5),(3,4),(4,3),(5,2),(6,1) — six outcomes.

6/36 = 1/6
Question 2

(a)P(1st red) = 5/8. After removing one red, P(2nd red) = 4/7. Multiply, since the events are dependent.

(5/8) × (4/7) = 20/56 = 5/14
Question 3

(a)With replacement, the two draws are independent: P(red) each time = 5/8.

(5/8) × (5/8) = 25/64
Question 4

(a)P(at least one head) = 1 − P(no heads) = 1 − P(tail, tail).

1 − (1/2×1/2) = 1 − 1/4 = 3/4
Question 5

(a)There are 6 matching pairs (1,1),(2,2)…(6,6) out of 36 total outcomes.

6/36 = 1/6
Question 6

(a)Without replacement: P(1st green)=6/10, P(2nd green | 1st green)=5/9, multiply.

(6/10)×(5/9) = 30/90 = 1/3
Question 7

(a)For independent events, multiply the probabilities.

0.7 × 0.7 = 0.49
Question 8

(a)Exactly one passing can happen two ways (1st passes & 2nd fails, OR 1st fails & 2nd passes): 2 × 0.7 × 0.3.

2 × 0.7 × 0.3 = 0.42
Question 9

(a)A 6×6 grid listing every combination of the two dice scores.

36 equally likely outcomes
Question 10

(a)List the outcomes summing to less than 5: (1,1),(1,2),(2,1),(1,3),(3,1),(2,2) — six outcomes out of 36.

6/36 = 1/6

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