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NIMCET 2013 — Mathematics PYQ

NIMCET | Mathematics | 2013

An experiment succeeds twice as often as it fails. The probability that in the next six trials there will be at least four successes, is:

Choose the correct answer:

Explanation

Solution

Step 1: Find the probability of success ( $p$ ) and failure ( $q$ )

Let the probability of failure be $q$ and the probability of success be $p$ .

According to the question, the experiment succeeds twice as often as it fails:

$p = 2q$

We know that $p + q = 1$ . Substituting $p = 2q$ :

$2q + q = 1 \implies 3q = 1 \implies q = \frac{1}{3}$

Therefore, $p = 2 \times \frac{1}{3} = \frac{2}{3}$ .

Step 2: Identify parameters for Binomial Distribution

Number of trials ( $n$ ) = $6$

Probability of success ( $p$ ) = $\frac{2}{3}$

Probability of failure ( $q$ ) = $\frac{1}{3}$

We need to find the probability of at least four successes ( $X \ge 4$ ):

$P(X \ge 4) = P(X = 4) + P(X = 5) + P(X = 6)$

Step 3: Calculate individual probabilities

Using the Binomial formula $P(X = r) = \binom{n}{r} p^r q^{n-r}$ :

For $r = 4$ :

$P(X = 4) = \binom{6}{4} \left(\frac{2}{3}\right)^4 \left(\frac{1}{3}\right)^2 = 15 \times \frac{16}{81} \times \frac{1}{9} = \frac{240}{729}$

For $r = 5$ :

$P(X = 5) = \binom{6}{5} \left(\frac{2}{3}\right)^5 \left(\frac{1}{3}\right)^1 = 6 \times \frac{32}{243} \times \frac{1}{3} = \frac{192}{729}$

For $r = 6$ :

$P(X = 6) = \binom{6}{6} \left(\frac{2}{3}\right)^6 \left(\frac{1}{3}\right)^0 = 1 \times \frac{64}{729} \times 1 = \frac{64}{729}$

Step 4: Sum the probabilities

$P(X \ge 4) = \frac{240}{729} + \frac{192}{729} + \frac{64}{729}$

$P(X \ge 4) = \frac{496}{729}$

Final Answer: The probability is $\frac{496}{729}$ .

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