Question

In a binomial distribution n = 6, p(3): p(4) = 8: 3 find the value
p.
​

Answer 1

SOLUTION

GIVEN

In a binomial distribution

n = 6, P(3) : P(4) = 8 : 3

TO DETERMINE

The value of p

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

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

EVALUATION

Here it is given that in a binomial distribution

n = 6, P(3) : P(4) = 8 : 3

So by the given condition -

P(3) : P(4) = 8 : 3

$\displaystyle \sf{ \implies \frac{ {}^{6} C_3\: {p}^{3} {(1 - p)}^{6 - 3} }{ {}^{6} C_4\: {p}^{4} {(1 - p)}^{6 - 4}} = \frac{8}{3} }$

$\displaystyle \sf{ \implies \frac{ 20\: {p}^{3} {(1 - p)}^{3} }{ 15\: {p}^{4} {(1 - p)}^{2}} = \frac{8}{3} }$

$\displaystyle \sf{ \implies \frac{ 4\:(1 - p) }{ 3\: p} = \frac{8}{3} }$

$\displaystyle \sf{ \implies \frac{ (1 - p) }{ p} = \frac{2}{1} }$

$\displaystyle \sf{ \implies 2p = (1 - p) }$

$\displaystyle \sf{ \implies 3p = 1}$

$\displaystyle \sf{ \implies p = \frac{1}{3} }$

FINAL ANSWER

$\boxed{ \: \: \displaystyle \sf{ p = \frac{1}{3} } \: \: }$

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