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14.Probability
normal
It is $5 : 2$ against a husband who is $65$ years old living till he is $85$ and $4 : 3$ against his wife who is now $58$, living till she is $78$. If the probability that atleast one of them will be alive for $20$ years, is $'k'$, then the value of $'49k'$ -
A
$20$
B
$31$
C
$29$
D
$6$
Solution
$\frac{\overline{\mathrm{H}}}{\mathrm{H}}=\frac{5}{2} ; \frac{\overline{\mathrm{w}}}{\mathrm{w}}=\frac{4}{3}$
$\mathrm{H}=\frac{2}{7}, \mathrm{w}=\frac{3}{7}$
$\overline{\mathrm{H}}=\frac{5}{7}, \overline{\mathrm{w}}=\frac{4}{7}$
$\Rightarrow 1-\overline{\mathrm{H}} \overline{\mathrm{w}}=\mathrm{K}$
$1-\frac{20}{49}=\mathrm{K}$
$49 \mathrm{K}=29$
Standard 11
Mathematics
