A die is thrown 4 times . Getting a 1 or a 6' is considered a success . Find probability of getting .
(i) exactly 3 successes (ii) at least 2 successes
(ii) at most 2 successes .

Explanation

To solve the given problem, we will use the concept of binomial probability distribution. Here’s a step-by-step breakdown of the solution:

### Step 1: Define the Problem
A die is thrown 4 times, and getting a 1 or a 6 is considered a success. We need to find the probability of:
1. Exactly 3 successes
2. At least 2 successes
3. At most 2 successes

### Step 2: Determine the Probability of Success and Failure
- The probability of success (getting a 1 or a 6) when throwing a die is:
\[ P(\text{Success}) = \frac{2}{6} = \frac{1}{3} \]
- The probability of failure (getting a 2, 3, 4, or 5) is:
\[ P(\text{Failure}) = 1 - P(\text{Success}) = 1 - \frac{1}{3} = \frac{2}{3} \]

### Step 3: Use the Binomial Distribution Formula
The binomial distribution formula is given by:
\[ P(X = r) = \binom{n}{r} p^r q^{n-r} \]
where:
- \( n \) = number of trials (4 in this case)
- \( r \) = number of successes
- \( p \) = probability of success (\(\frac{1}{3}\))
- \( q \) = probability of failure (\(\frac{2}{3}\))

### Part (i): Probability of Exactly 3 Successes
- Here, \( r = 3 \):
\[ P(X = 3) = \binom{4}{3} \left(\frac{1}{3}\right)^3 \left(\frac{2}{3}\right)^{4-3} \]

Calculating:
\[ \binom{4}{3} = 4 \]
\[ P(X = 3) = 4 \left(\frac{1}{3}\right)^3 \left(\frac{2}{3}\right)^1 = 4 \cdot \frac{1}{27} \cdot \frac{2}{3} = 4 \cdot \frac{2}{81} = \frac{8}{81} \]

### Part (ii): Probability of At Least 2 Successes
- At least 2 successes means \( r = 2, 3, \) or \( 4 \). We can calculate this by finding the complement (0 or 1 success):
\[ P(X \geq 2) = 1 - P(X = 0) - P(X = 1) \]

Calculating \( P(X = 0) \):
\[ P(X = 0) = \binom{4}{0} \left(\frac{1}{3}\right)^0 \left(\frac{2}{3}\right)^4 = 1 \cdot 1 \cdot \left(\frac{16}{81}\right) = \frac{16}{81} \]

Calculating \( P(X = 1) \):
\[ P(X = 1) = \binom{4}{1} \left(\frac{1}{3}\right)^1 \left(\frac{2}{3}\right)^3 = 4 \cdot \frac{1}{3} \cdot \frac{8}{27} = \frac{32}{81} \]

Now, putting it all together:
\[ P(X \geq 2) = 1 - \left(\frac{16}{81} + \frac{32}{81}\right) = 1 - \frac{48}{81} = \frac{33}{81} = \frac{11}{27} \]

### Part (iii): Probability of At Most 2 Successes
- At most 2 successes means \( r = 0, 1, \) or \( 2 \):
\[ P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) \]

We already calculated \( P(X = 0) \) and \( P(X = 1) \). Now we calculate \( P(X = 2) \):
\[ P(X = 2) = \binom{4}{2} \left(\frac{1}{3}\right)^2 \left(\frac{2}{3}\right)^2 = 6 \cdot \frac{1}{9} \cdot \frac{4}{9} = \frac{24}{81} \]

Now, summing them up:
\[ P(X \leq 2) = \frac{16}{81} + \frac{32}{81} + \frac{24}{81} = \frac{72}{81} = \frac{8}{9} \]

### Final Answers:
1. Probability of exactly 3 successes: \(\frac{8}{81}\)
2. Probability of at least 2 successes: \(\frac{11}{27}\)
3. Probability of at most 2 successes: \(\frac{8}{9}\)