A die is thrown 6 times. If “getting an odd number” is a “success”, what is the probability of (i) 5 successes ? (ii) atmost 5 successes ? 

 

 

 

 
 
 
 
 

Answers (1)

Given n=6

Success = Getting an odd number  =\left \{ 1,3,5 \right \}

p : probablity of success

q : probablity of failure 

p=\frac{3}{6}=\frac{1}{2},  q=\frac{3}{6}=\frac{1}{2}

(i)  P(5\: \: successes)=\!^{n}\textrm{C}_5p^5q^1

                                        =\! ^6\textrm{C}_{5} \left ( \frac{1}{2} \right )^5\! \left ( \frac{1}{2} \right )^1

                                                                                                                          =6 \times \frac{1}{32}\times \frac{1}{2}=\frac{6}{64}=\frac{3}{32}

(ii) P(atmost\: \: successes)=P(x\leq 5)

                                                    =1-P(x=6)

                                                     =1-\!^{6}\textrm{C}_6 \left ( \frac{1}{2} \right )^6

                                                     =1-\frac{1}{64}=\frac{63}{64}

                                        

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