Question

In a trial, the probability of success is twice the probability of failure. In six trials, the probability of at most two failures will be

• A. 600/729
• B. 500/729
• C. 400/729
• D. 496/729

Hint:

To solve probability problems involving binomial distributions, always identify the number of trials, the probability of success and failure, and apply the binomial probability formula for each case of interest.

Answer 1

The Correct Option is D

Let the probability of success be \( P(S) \) and the probability of failure be \( P(F) \). 
According to the given information, the probability of success is twice the probability of failure, so: \[ P(S) = 2P(F) \] Since the total probability is 1, we can express this as: \[ P(S) + P(F) = 1 \] 
Substituting \( P(S) = 2P(F) \) into the equation: \[ 2P(F) + P(F) = 1 \implies 3P(F) = 1 \implies P(F) = \frac{1}{3} \] Thus, \( P(S) = 2 \times \frac{1}{3} = \frac{2}{3} \). 
Now, the probability of at most two failures in six trials is the sum of the probabilities of having 0, 1, or 2 failures. 
This is a binomial probability problem, so we use the binomial distribution formula: \[ P(k \text{ failures}) = \binom{n}{k} P(F)^k P(S)^{n-k} \] For \( n = 6 \) trials, the probability of at most two failures is: \[ P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) \] 
We calculate each term: 

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

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

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

Thus, the total probability of at most two failures is: \[ P(X \leq 2) = \frac{64}{729} + \frac{192}{729} + \frac{240}{729} = \frac{496}{729} \]

Answer 2

Let the probability of success be \( p \) and the probability of failure be \( q \). We know:

• The probability of success is twice the probability of failure: \( p = 2q \).
• The total probability must equal 1: \( p + q = 1 \).

Substituting the first equation into the second, we have:

\( 2q + q = 1 \\ 3q = 1 \\ q = \frac{1}{3} \)

Thus, the probability of success, \( p \), is:

\( p = 2q = 2 \times \frac{1}{3} = \frac{2}{3} \)

In six trials, we need the probability of at most 2 failures. Denote this as \( P(X \leq 2) \) where \( X \) is the number of failures, which follows a binomial distribution \( X \sim \text{Binomial}(n=6, q=\frac{1}{3}) \).

The probability \( P(X \leq 2) \) is:

\( P(X = 0) + P(X = 1) + P(X = 2) \)

Using the binomial probability formula:

\( P(X = k) = \binom{n}{k} (q)^k (p)^{n-k} \)

Calculate each:

• \( P(X = 0) = \binom{6}{0} \left(\frac{1}{3}\right)^0 \left(\frac{2}{3}\right)^6 = 1 \times 1 \times \left(\frac{64}{729}\right) = \frac{64}{729} \)
• \( P(X = 1) = \binom{6}{1} \left(\frac{1}{3}\right)^1 \left(\frac{2}{3}\right)^5 = 6 \times \frac{1}{3} \times \frac{32}{243} = \frac{192}{729} \)
• \( P(X = 2) = \binom{6}{2} \left(\frac{1}{3}\right)^2 \left(\frac{2}{3}\right)^4 = 15 \times \frac{1}{9} \times \frac{16}{81} = \frac{240}{729} \)

Add them up:

\( P(X \leq 2) = \frac{64}{729} + \frac{192}{729} + \frac{240}{729} = \frac{496}{729} \)

Thus, the probability of at most two failures is 496/729.