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A cook has 10 red peppers and 5 green peppers. If the cook selects 6 peppers at random, what is the probability that he selects at least \(4\) green peppers? Express your answer as a common fraction.

 Apr 16, 2022
 #1
avatar+124594 
+2

P (4 green peppers)  =   C(5,4) * C (10,2)  / C(15 ,6)    = 45  /1001

P(5 green pepers)   = C(5,5) * C(10,1)  /C(15,6)    =  2 /1001

 

Total probability   =  [45 + 2]  /1001  =   47 /1001

 

cool cool cool

 Apr 16, 2022
 #2
avatar+9459 
+1

Note that he can choose (i) 4 green peppers and 2 red peppers, or (ii) 5 green peppers and 1 red pepper. Then,

\(\begin{array}{rcl} \text{Probability} &=&\dfrac{\displaystyle \binom{5}4\binom{10}2+\binom{5}5\binom{10}1}{\displaystyle \binom{10+5}6}\\ &=& \dfrac{47}{1001} \end{array}\)

 Apr 16, 2022

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