Test Index

Probability Part 1

Section: Mathematics

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Question : 84 of 100

Marks: +1, -0

In a binomial distribution

B \left(n, p=\frac{1}{4}\right)

, if the probability of at least one success is greater than or equal to

\; \frac{9}{10}

n

is greater than:

$\; \frac{1}{\log_{10}^4+\log_{10}^3}$
$\; \frac{9}{\log_{10}^4-\log_{10}^3}$
$\; \frac{4}{\log_{10}^4-\log_{10}^3}$
$\; \frac{1}{\log_{10}^4-\log_{10}^3}$

Solution:

Given that, no of trials

=n

Probability of success

(p)=\; \frac{1}{4}

\therefore

Probability of no success

=1-\; \frac{1}{4}=\; \frac{3}{4}

As we know, probability of at least one success

=1

- probability of no success

\therefore

According to the question,

1 - (Probability of no success in

n

trials

) \geq \; \frac{9}{10}

\Rightarrow 1- P(x=0) \geq \; \frac{9}{10}

\Rightarrow 1- {}^n C_0 \left(\; \frac{1}{4}\right)^0 \left(\; \frac{3}{4}\right)^n \geq \; \frac{9}{10}

\Rightarrow 1- \left(\; \frac{3}{4}\right)^n \geq \; \frac{9}{10}

\Rightarrow \left(\; \frac{3}{4}\right)^n \leq \; \frac{1}{10}

\Rightarrow \left(\; \frac{4}{3}\right)^n \geq 10

[Taking log on both sides]

\; \Rightarrow n (\log_{10}^4-\log_{10}^3) \geq \log_{10}^{10}

\; \Rightarrow n (\log_{10}^4-\log_{10}^3) \geq 1

\; \Rightarrow n \geq \; \frac{1}{(\log_{10}^4-\log_{10}^3)}

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