Solution the probability that a patient recovers from a rare blood disease is 0.4. If 15 people are known to have contacted this disease,what is the probability that (a) at least 10 survive? (b) from 3 to 8 survive? (c) exactly 5 survive
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Answer:
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Given : probability that a patient recovers from a rare blood disease is 0.4 , 15 people are known to have contacted this disease
To find : probability that (a) at least 10 survive? (b) from 3 to 8 survive? (c) exactly 5 survive
Solution:
probability that a patient recovers from a rare blood disease is 0.4
=> p = 0.4
Does not recover q = 1 - 0.4 = 0.6
people = n = 15
p(x) = ⁿCₓpˣqⁿ⁻ˣ
probability that (a) at least 10 survive
10 ,11 , 12 , 13 ,1 4 , 15 survive
= p(10) + p(11) + p(12) + p(13) + p(14) + p(15)
= ¹⁵C₁₀(0.4)¹⁰(0.6)⁵ + ¹⁵C₁₁(0.4)¹¹(0.6)⁴ + ¹⁵C₁₂(0.4)¹²(0.6)³ + ¹⁵C₁₃(0.4)¹³(0.6)² + ¹⁵C₁₄(0.4)¹⁴(0.6)¹ + ¹⁵C₁₅(0.4)¹⁵(0.6)⁰
(b) from 3 to 8 survive
p(3) + p(4) + p(5) + p(6) + p(7) + p(8)
= ¹⁵C₃(0.4)³(0.6)¹² + ¹⁵C₄(0.4)⁴(0.6)¹¹ + ¹⁵C₅(0.4)⁵(0.6)¹⁰ + ¹⁵C₆(0.4)⁶(0.6)⁹ + ¹⁵C₇(0.4)⁷(0.6)⁸ + ¹⁵C₈(0.4)⁸(0.6)⁷
(c) exactly 5 survive
= ¹⁵C₅(0.4)⁵(0.6)¹⁰
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