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  • A box contains 4 red balls and 6 black balls. Three balls are selected randomly from the box one after another, without replacement. The probability that the selected set contains one red ball and

A box contains 4 red balls and 6 black balls. Three balls are selected randomly from the box one after another, without replacement. The probability that the selected set contains one red ball and two black balls is
 

Option: 1

\frac{1}{20}
 


Option: 2

\frac{1}{12}
 


Option: 3

\frac{3}{10}
 


Option: 4

\frac{1}{2}


Answers (1)

best_answer

Given, a box contains 4 red balls $\& 6$ black balls. Then, probability that three balls selected contains are red ball \& two black balls, without replacement is

\mathrm{ P_T=P(R B B)+P(B R B)+P(B B R) }

\mathrm{P(R) \rightarrow} Probability of red ball selection

\mathrm{P(B) \rightarrow}Probability of black ball selection

\mathrm{ P_T=\frac{4}{10} \times \frac{6}{9} \times \frac{5}{8}+\frac{6}{10} \times \frac{4}{9} \times \frac{5}{8}+\frac{6}{10} \times \frac{5}{9} \times \frac{4}{8} }

\mathrm{ \Rightarrow \quad P_T=\frac{1}{2} }
 

Posted by

Gaurav

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