Question

In a binomial distribution , p=1/2,q=1/2, n =6. Then p(x=2)is

Answer 1

$\underline{\textsf{Given:}}$

$\textsf{In a binomial distribution,}$

$\mathsf{n=6,\;p=\dfrac{1}{2},\;q=\dfrac{1}{2}}$

$\underline{\textsf{To find:}}$

$\mathsf{P(X=2)}$

$\underline{\textsf{Solution:}}$

$\textsf{Concept used:}$

$\textsf{The probability mass function of binomial distribution is}$

$\mathsf{P(X=x)=\,_nC_x\;p^x\;q^{n-x}\;\;x=0,1,2.......n}$

$\textsf{Now}$

$\mathsf{P(X=2)}$

$\mathsf{=\,_6C_2\,(\dfrac{1}{2})^2\,(\dfrac{1}{2})^4}$

$\mathsf{=\dfrac{6{\times}5}{1{\times}2}\,(\dfrac{1}{2})^6}$

$\mathsf{=3{\times}5\,(\dfrac{1}{2})^6}$

$\mathsf{=\dfrac{15}{64}}$

$\implies\boxed{\mathsf{P(X=2)=\dfrac{15}{64}}}$

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

GIVEN

$\displaystyle \sf{ In \: a \: binomial \:distribution \: \: p = \frac{1}{2} \: , \: q = \frac{1}{2} \: , \: n = 6 }$

TO DETERMINE

$\sf{P(X=2)}$

FORMULA TO BE IMPLEMENTED

If a trial is repeated n times and p is the probability of a success and q that of failure then the probability of r successes is

$\displaystyle \sf{ \sf{P(X=r) = \: \: }\large{ {}^{n} C_r}\: {p}^{r} \: \: {q}^{n - r} } \: \: \: \: \: where \: q \: = 1 - p$

CALCULATION

It is given that

$\displaystyle \sf{ In \: a \: binomial \:distribution \: \: p = \frac{1}{2} \: , \: q = \frac{1}{2} \: , \: n = 6 }$

Hence

$\displaystyle \sf{ \sf{P(X=2) = \: \: }\large{ {}^{6} C_2}\: {p}^{2} \: \: {q}^{6 - 2} } \: \:$

$\displaystyle \sf{ = \large{ {}^{6} C_2}\: { \bigg( \frac{1}{2} \bigg)}^{2} \: \: { \bigg(1 - \frac{1}{2} \bigg)}^{6 - 2} } \: \:$

$\displaystyle \sf{ = \frac{6 \:! }{2 \: ! \: \times \: 4 \:! } \: \: { \bigg( \frac{1}{2} \bigg)}^{2} \: \: { \bigg(1 - \frac{1}{2} \bigg)}^{6 - 2} } \: \:$

$\displaystyle \sf{ = \frac{6 \times 5}{2} \: \times { \bigg( \frac{1}{2} \bigg)}^{2} \: \: { \bigg( \frac{1}{2} \bigg)}^{4} } \: \:$

$\displaystyle \sf{ = 15 \times { \bigg( \frac{1}{2} \bigg)}^{6} \: \: } \: \:$

$\displaystyle \sf{ = \frac{15}{64} } \: \:$

RESULT

$\boxed{ \: \: \: \sf{P(X=2)} \displaystyle \sf{ \: \: \: = \: \: \frac{15}{64} } \: \: \: }$

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