If a biased coin is tossed 5 times and if the probability getting 4 head is equal to the probability of geeting 5 heads , then the probability of getting at most 2 head is

Options

A. `46/6^4`
B. `275/6^5`
C. `51/5^5`
D. `36/5^4`

Answer

To solve the problem, we will follow these steps:

### Step 1: Define the probabilities
Let \( P \) be the probability of getting heads, and \( Q \) be the probability of getting tails. Since the coin is biased, we have:
\[ P + Q = 1 \]

### Step 2: Set up the equations based on the problem statement
According to the problem, the probability of getting 4 heads is equal to the probability of getting 5 heads when the coin is tossed 5 times. We can express these probabilities using the binomial distribution formula:

- Probability of getting 4 heads:
\[ P(X = 4) = \binom{5}{4} P^4 Q^1 = 5 P^4 Q \]

- Probability of getting 5 heads:
\[ P(X = 5) = \binom{5}{5} P^5 Q^0 = P^5 \]

Setting these two probabilities equal gives us:
\[ 5 P^4 Q = P^5 \]

### Step 3: Simplify the equation
We can simplify this equation by dividing both sides by \( P^4 \) (assuming \( P \neq 0 \)):
\[ 5 Q = P \]

### Step 4: Use the relationship between \( P \) and \( Q \)
From the first equation \( P + Q = 1 \), we can substitute \( P \) with \( 5Q \):
\[ 5Q + Q = 1 \implies 6Q = 1 \implies Q = \frac{1}{6} \]
Now, substituting back to find \( P \):
\[ P = 5Q = 5 \cdot \frac{1}{6} = \frac{5}{6} \]

### Step 5: Calculate the probability of getting at most 2 heads
We need to find the probability of getting at most 2 heads, which means we need to calculate \( P(X \leq 2) \):
\[ P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) \]

Calculating each term:
- For \( X = 0 \):
\[ P(X = 0) = \binom{5}{0} P^0 Q^5 = 1 \cdot 1 \cdot \left(\frac{1}{6}\right)^5 = \frac{1}{6^5} \]

- For \( X = 1 \):
\[ P(X = 1) = \binom{5}{1} P^1 Q^4 = 5 \cdot \frac{5}{6} \cdot \left(\frac{1}{6}\right)^4 = 5 \cdot \frac{5}{6} \cdot \frac{1}{6^4} = \frac{25}{6^5} \]

- For \( X = 2 \):
\[ P(X = 2) = \binom{5}{2} P^2 Q^3 = 10 \cdot \left(\frac{5}{6}\right)^2 \cdot \left(\frac{1}{6}\right)^3 = 10 \cdot \frac{25}{36} \cdot \frac{1}{216} = \frac{250}{6^5} \]

### Step 6: Combine the probabilities
Now, we can sum these probabilities:
\[ P(X \leq 2) = \frac{1}{6^5} + \frac{25}{6^5} + \frac{250}{6^5} = \frac{1 + 25 + 250}{6^5} = \frac{276}{6^5} \]

### Final Answer
Thus, the probability of getting at most 2 heads is:
\[ \frac{276}{6^5} \]