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  • If a random variable follows the Binomial distribution <img alt="\mathrm{B(33, p)}" src="https://entr

If a random variable \mathrm{X} follows the Binomial distribution \mathrm{B(33, p)} such that \mathrm{3 P(X=0)=P(X=1)}, then the value of \mathrm{\frac{\mathrm{P}(\mathrm{X}=15)}{\mathrm{P}(\mathrm{X}=18)}-\frac{\mathrm{P}(\mathrm{X}=16)}{\mathrm{P}(\mathrm{X}=17)}}   is equal to:

Option: 1

\mathrm{1320}


Option: 2

\mathrm{1088}


Option: 3

\mathrm{\frac{120}{1331}}


Option: 4

\mathrm{\frac{1088}{1089}}


Answers (1)

best_answer

\mathrm{3P\left ( X= 0 \right )= P\left ( X= 1 \right )}
\mathrm{3\cdot \, ^{33}C_{0}\, \left ( q \right )^{33}= \, ^{33}C_{1}\, p\left ( q \right )^{32}}
\mathrm{q^{33}=11\cdot p\, q^{32}}
\mathrm{\frac{q}{P}= 11}

As p + q = 1
\mathrm{\therefore \: \: P= \frac{1}{12},\: q= \frac{11}{12}}

\therefore \: \mathrm{\frac{\mathrm{P}(\mathrm{X}=15)}{\mathrm{P}(\mathrm{X}=18)}-\frac{\mathrm{P}(\mathrm{X}=16)}{\mathrm{P}(\mathrm{X}=17)}}

\mathrm{\frac{^{33}C_{15}\, \, p^{15} \,\, q^{18}}{^{33}C_{18}\, \, p^{18} \,\, q^{15}}-\frac{^{33}C_{16}\, \, p^{16} \,\, q^{17}}{^{33}C_{17}\, \, p^{17} \,\, q^{16}}}

\mathrm{\left ( \frac{q}{p} \right )^{3}-\left ( \frac{q}{p} \right )= 1331-11}
                             \mathrm{= 1320}

The correct answer is option (A)

Posted by

Shailly goel

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