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The probability that a student is not a swimmer is 1 / 5. Then the probability that out of five students, four are swimmers is?

\[ \begin{array} ((A) \: {C}_4^5 \bigg(\frac{4}{5} \bigg)^4 \: \frac{1}{5} \quad & (B)\; \bigg(\frac{4}{5}\bigg)^4 \: \frac{1}{5} \\ (C)\: {C}_1^5 \: \frac{1}{5} \: \bigg(\frac{4}{5}\bigg)^4 \quad & (D)\: None \: of\: these \end{array} \]

1 Answer

Toolbox:
  • For any Binomial distribution $B (n, p),$ the probability of x success in n-Bernoulli trials, $P (X = x) = \large^{n}C_x. p^x.q^{n–x}$ where $x = 0, 1, 2,...,n$ and $(q = 1 – p)$
  • Trials of a random experiment are called Bernoulli trials, if they satisfy the following conditions: (i) There should be a finite number of trials. (ii) The trials should be independent. (iii) Each trial has exactly two outcomes : success or failure. (iv) The probability of success remains the same in each trial.
Let X be the number of students out of $ (n=5)$ students that is a swimmer.
It is a Bernoulli trial as they satisfy the conditions (i) finite number of trials, (ii) independent trials, (iii) there is a definite outcome and (iv) the probability of success does not change for each trial..
P (not a swimmer) $q = \large\frac{1}{5} \rightarrow $$ p = 1-q = \large\frac{4}{5}$
Since X has a bionomial distribution, the probability of x success in n-Bernoulli trials, $P (X = x) = \large^{n}C_x. p^x.q^{n–x}$ where $x = 0, 1, 2,...,n$ and $(q = 1 – p)$
P (four students are swimmer), $ P (X = 4) = \large^{5}C_4\large(\frac{4}{5})^4\large(\frac{1}{5})^{5-4}$
$P (X = 4) = \large^{5}C_4\large(\frac{4}{5})^4\large(\frac{1}{5})$
answered Jun 21, 2013 by balaji.thirumalai
 

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