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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