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Out of 11 consecutive natural no if 3 are selected at random without repeat Then the probability that they are in AP with +ve c.d  
 
Option: 1 \frac{5}{33}
Option: 2 \frac{10}{99}
Option: 3 \frac{3}{101}  
Option: 4 \frac{15}{101}

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\begin{aligned} \text { Total no of selections } &=\begin{array}{c} ^{11}C_3 \end{array} \\ &=\frac{11 \times 10 \times 9}{6} \\ &=165 \end{aligned}

No of ways with common difference 1 = 9

No of ways with common difference 2 = 7

No of ways with common difference 3 = 5

No of ways with common difference 4 = 3

No of ways with common difference 5 = 1

\\ \text { Required probability }=\frac{1+3+5+7+9}{165} \\ \qquad \begin{aligned} &=\frac{25}{165}=\frac{5}{33} \end{aligned}

Correct Answer: Option A

Posted by

himanshu.meshram

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