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  • 3 June, 2026 at 8:24 pm #1113

    Two distinct numbers a and b are selected at random from
    $1,2,3, \ldots \ldots, 50$. The probability, that their product ab is divisible by 3 , is
    [JEE MAIN 22 Jan 2026 S I]
    (1) $\frac{561}{1225}$
    (2) $\frac{664}{1225}$
    (3) $\frac{272}{1225}$
    (4) $\frac{8}{25}$

    3 June, 2026 at 8:24 pm #1114

    Solution

    First Method:

    $\mathrm{P}(\mathrm{ab}$ is divisible by 3$)$
    $=\mathrm{P}(\mathrm{a} \& \mathrm{~b}$ both are multiple of 3$)+\mathrm{P}($ Exactly are of $\mathrm{a} \& \mathrm{~b}$ is multiple of 3$)$
    $=\frac{{ }^{16} \mathrm{C}_{2}}{{ }^{50} \mathrm{C}_{2}}+\frac{{ }^{16} \mathrm{C}_{1} \times{ }^{34} \mathrm{C}_{1}}{{ }^{50} \mathrm{C}_{2}}=\frac{8.15+16.34}{25(49)}=\frac{664}{1225}$
    $3 \mathrm{~K}=\{3,6,9, \ldots, 48\}$

    Second Method:

    Number which are not multiple of $3=50-16=34$
    $\mathrm{P}(\mathrm{ab}$ is divisible by 3$)=1-\mathrm{P}(\mathrm{ab}$ is not divisible by 3$)$
    $=1-\frac{{ }^{34} \mathrm{C}_{2}}{{ }^{50} \mathrm{C}_{2}}=1-\frac{17 \times 93}{25 \times 49}=\frac{1225-561}{1225}=\frac{664}{1225}$ Answer(2)

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