QUESTION 5 A cricket club has 15 members, of whom only 5 can bowl.
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QUESTION 5
A cricket club has 15 members, of whom only 5 can bowl. If the names of the 15 members are put into a box and 11 drawn at random, then the chance of obtaining at least 3 bowlers is:
Show answer & explanation
When drawing 11 members from 15 (5 bowlers, 10 non-bowlers), we use the complement: P(at least 3 bowlers) = 1 - P(fewer than 3 bowlers) = 1 - P(0, 1, or 2 bowlers). The only ways to get fewer than 3 bowlers are: (1) all 11 drawn are non-bowlers (impossible—only 10 non-bowlers exist) or (2) exactly 1 bowler and 10 non-bowlers. The probability of getting exactly 1 bowler is C(5,1)×C(10,10) / C(15,11) = 5/1001. Thus P(at least 3) = 1 - 5/1001 = 996/1001 ≈ 12/13.
Step-by-step Derivation:
Total members: 15 (5 bowlers, 10 non-bowlers). Drawing 11 at random.
Total ways to choose 11 from 15: C(15,11) = C(15,4) = (15×14×13×12)/(4×3×2×1) = 1365
P(at least 3 bowlers) = 1 - P(fewer than 3 bowlers)
Fewer than 3 bowlers means 0, 1, or 2 bowlers:
- P(0 bowlers) = C(5,0)×C(10,11) / C(15,11) = 0 (impossible, only 10 non-bowlers)
- P(1 bowler) = C(5,1)×C(10,10) / C(15,11) = (5×1) / 1365 = 5/1365 = 1/273
- P(2 bowlers) = C(5,2)×C(10,9) / C(15,11) = (10×10) / 1365 = 100/1365 = 20/273
P(fewer than 3) = 1/273 + 20/273 = 21/273 = 1/13
P(at least 3 bowlers) = 1 - 1/13 = 12/13