Question

The probability that a patient suffering from spontaneous pneumothorax (sudden collapsed lung) will require collective surgery is 0.6. What is the probability that 3 to 10 of the next 15 patients will require the surgery? Answer in decimal form.

Answer 1

Let X be the number of next 15 patients suffering from spontaneous pneumothorax require collective surgery

X ~ Binomial(n = 15, p = 0.6)

P(3 $\le$ X $\le$ 10) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)

$= \binom{15}{3} 0.6^3 * 0.4^{12} + \binom{15}{4} 0.6^4 * 0.4^{11} + \binom{15}{5} 0.6^5 * 0.4^{10} + \binom{15}{6} 0.6^6 * 0.4^{9} +$

(1)0.6++0.4° +(6)0.8 +0.4² + (3)0.6° +0.4°+ (10)6.6 40 *0.45

= 0.001648865 + 0.007419892 + 0.024485642 + 0.061214105 + 0.118055774 + 0.177083662 + 0.206597605 + 0.185937845

= 0.7824434