Question 1130144
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(a)  P(2) = {{{C[5]^2*0.3^2*(1-0.3)^3}}} = {{{((5*4)/(1*2))*0.3^2*(1-0.3)^3}}} = 0.3087,  exactly as you got (!)



(b)  "catches at least 2" means "catches 2, or 3, or 4, or 5" 


     P = P(2) + P(3) + P(4) + P(5) = (which is  <U>EASIER</U>  to calculate) = 1 - P(0) - P(1)  = 1 - {{{0.7^5}}} - {{{5*0.3*0.7^4}}} = 0.4718.
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