A die is thrown 6 times. If "getting an even number" is a success, what is the probability of (i) 5 successes? (ii) at least 5 successes? (iii) at most 5 successes?

Explanation

To solve the problem, we will use the binomial probability formula. The probability of getting exactly k successes in n trials is given by:

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

where:
- \( n \) = number of trials (in this case, 6)
- \( k \) = number of successes
- \( p \) = probability of success on a single trial
- \( q \) = probability of failure on a single trial

### Step 1: Determine the probabilities of success and failure

In a single throw of a die, the even numbers are 2, 4, and 6. Therefore, the probability of getting an even number (success) is:

\[ p = \frac{3}{6} = \frac{1}{2} \]

The probability of not getting an even number (failure) is:

\[ q = 1 - p = 1 - \frac{1}{2} = \frac{1}{2} \]

### Step 2: Calculate the probability of getting exactly 5 successes

We need to find \( P(X = 5) \):

\[ P(X = 5) = \binom{6}{5} p^5 q^{6-5} \]

Calculating \( \binom{6}{5} \):

\[ \binom{6}{5} = 6 \]

Now substituting the values of \( p \) and \( q \):

\[ P(X = 5) = 6 \left(\frac{1}{2}\right)^5 \left(\frac{1}{2}\right)^1 = 6 \left(\frac{1}{2}\right)^6 = 6 \cdot \frac{1}{64} = \frac{6}{64} = \frac{3}{32} \]

### Step 3: Calculate the probability of getting at least 5 successes

To find \( P(X \geq 5) \), we need to consider both \( P(X = 5) \) and \( P(X = 6) \):

\[ P(X \geq 5) = P(X = 5) + P(X = 6) \]

We already calculated \( P(X = 5) \). Now we calculate \( P(X = 6) \):

\[ P(X = 6) = \binom{6}{6} p^6 q^{6-6} = 1 \cdot \left(\frac{1}{2}\right)^6 \cdot 1 = \frac{1}{64} \]

Now adding both probabilities:

\[ P(X \geq 5) = P(X = 5) + P(X = 6) = \frac{3}{32} + \frac{1}{64} \]

To add these fractions, we need a common denominator:

\[ \frac{3}{32} = \frac{6}{64} \]

So,

\[ P(X \geq 5) = \frac{6}{64} + \frac{1}{64} = \frac{7}{64} \]

### Step 4: Calculate the probability of getting at most 5 successes

To find \( P(X \leq 5) \), we can use the complement rule:

\[ P(X \leq 5) = 1 - P(X = 6) \]

We already calculated \( P(X = 6) \):

\[ P(X = 6) = \frac{1}{64} \]

Thus,

\[ P(X \leq 5) = 1 - \frac{1}{64} = \frac{64 - 1}{64} = \frac{63}{64} \]

### Final Answers:

(i) The probability of getting exactly 5 successes is \( \frac{3}{32} \).

(ii) The probability of getting at least 5 successes is \( \frac{7}{64} \).

(iii) The probability of getting at most 5 successes is \( \frac{63}{64} \).