Question

A new surgery is successful 70% of the time. If the results of 7 such surgeries are randomly sampled, what is the probability that at most 4 of them are Successful? Carry your intermediate computations to at least four decimal places, and round your answer to two decimal places (If necessary, consult a list of formulas.) 0 X ?

Answer 1

Solution :

Given that ,

p = 0.70

1 - p = 1 -0.70 = 0.30

n = 7

So,

p (x $\leq$ 4 ) =p (x = 0 )+ p (x = 1) + p (x = 2 ) +  p (x = 3 ) +  p (x = 4 )     

  = (7! / (7 - 0)!) * 0.70⁰ * 0.30)^{7 - 0}

  = (7! / (7 - 1)!) * 0.70 * 0.30)^{7- 1}

  = (7! / (7 - 2)!) * 0.70 * 0.30)^{7 - 2}

  = (7! / (7 - 3)!) * 0.70 * 0.30)^{7 - 3}

= (7! / (7 - 4)!) * 0.70 * 0.30)^{7 -4}

= 0.0002 + 0.0036 + 0.0250 +0.0972 +0.2269

= 0.3529

p ( x $\leq$ ​​​​​​​ 4 ) = 0.3529

Probability = 0.3529