Biased coins have probability of getting head is `2/3 and x` is number of heads when six coins are tossed , then the probability `P(X le 2)` is equal to

Options

A. `73/729`
B. `67/729`
C. `23/729`
D. `73/243`

Answer

To solve the problem of finding the probability \( P(X \leq 2) \) when tossing six biased coins with the probability of getting heads \( p = \frac{2}{3} \), we can use the binomial distribution. Here are the step-by-step calculations:

### Step 1: Define the parameters
We have:
- Number of trials (coin tosses), \( n = 6 \)
- Probability of success (getting heads), \( p = \frac{2}{3} \)
- Probability of failure (getting tails), \( q = 1 - p = \frac{1}{3} \)

### Step 2: Set up the binomial probability formula
The probability of getting exactly \( r \) heads in \( n \) tosses is given by the binomial probability formula:
\[ P(X = r) = \binom{n}{r} p^r q^{n-r} \]
where \( \binom{n}{r} \) is the binomial coefficient.

### Step 3: Calculate \( P(X = 0) \)
For \( r = 0 \):
\[ P(X = 0) = \binom{6}{0} \left(\frac{2}{3}\right)^0 \left(\frac{1}{3}\right)^{6} = 1 \cdot 1 \cdot \left(\frac{1}{729}\right) = \frac{1}{729} \]

### Step 4: Calculate \( P(X = 1) \)
For \( r = 1 \):
\[ P(X = 1) = \binom{6}{1} \left(\frac{2}{3}\right)^1 \left(\frac{1}{3}\right)^{5} = 6 \cdot \frac{2}{3} \cdot \left(\frac{1}{243}\right) = \frac{12}{729} \]

### Step 5: Calculate \( P(X = 2) \)
For \( r = 2 \):
\[ P(X = 2) = \binom{6}{2} \left(\frac{2}{3}\right)^2 \left(\frac{1}{3}\right)^{4} = 15 \cdot \left(\frac{4}{9}\right) \cdot \left(\frac{1}{81}\right) = \frac{60}{729} \]

### Step 6: Sum the probabilities
Now, we need to find \( P(X \leq 2) \):
\[ P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) \]
\[ P(X \leq 2) = \frac{1}{729} + \frac{12}{729} + \frac{60}{729} = \frac{73}{729} \]

### Final Answer
Thus, the probability \( P(X \leq 2) \) is:
\[ \boxed{\frac{73}{729}} \]