P(6)= (10C6) * (0.897)^6 * (0.103)^(10-6) P(7) = (10C7) * (0.897)^7 * (0.103)^(10-7) P(8) = (10C8) * (0.897)^8 * (0.103)^(10-8) P(9) = (10C9) * (0.897)^9 * (0.103)^(10-9) P(10) = (10C10) * (0.897)^10 * (0.103)^0=

Answer 1

To calculate the probabilities \( P(6) \), \( P(7) \), \( P(8) \), \( P(9) \), and \( P(10) \), we use the binomial probability formula:

\[ P(k) = \binom{n}{k} p^k (1-p)^{n-k} \]

where:

• \( n = 10 \) (the number of trials)
• \( p = 0.897 \) (the probability of success)
• \( (1-p) = 0.103 \) (the probability of failure)
• \( k \) is the number of successes.

We already have the formulas set up for \( P(6) \), \( P(7) \), \( P(8) \), \( P(9) \), and \( P(10) \).

Now we can calculate them one by one.

Probability \( P(6) \)

\[ P(6) = \binom{10}{6} (0.897)^6 (0.103)^{4} \]

Calculating the components:

• \( \binom{10}{6} = 210 \)
• \( (0.897)^6 \approx 0.536 \)
• \( (0.103)^4 \approx 0.000113 \)

Putting it all together:

\[ P(6) \approx 210 \cdot 0.536 \cdot 0.000113 \approx 0.01261 \]

Probability \( P(7) \)

\[ P(7) = \binom{10}{7} (0.897)^7 (0.103)^{3} \]

Calculating the components:

• \( \binom{10}{7} = 120 \)
• \( (0.897)^7 \approx 0.479 \)
• \( (0.103)^3 \approx 0.001092 \)

Putting it all together:

\[ P(7) \approx 120 \cdot 0.479 \cdot 0.001092 \approx 0.06307 \]

Probability \( P(8) \)

\[ P(8) = \binom{10}{8} (0.897)^8 (0.103)^{2} \]

Calculating the components:

• \( \binom{10}{8} = 45 \)
• \( (0.897)^8 \approx 0.429 \)
• \( (0.103)^2 \approx 0.010609 \)

Putting it all together:

\[ P(8) \approx 45 \cdot 0.429 \cdot 0.010609 \approx 0.20581 \]

Probability \( P(9) \)

\[ P(9) = \binom{10}{9} (0.897)^9 (0.103)^{1} \]

Calculating the components:

• \( \binom{10}{9} = 10 \)
• \( (0.897)^9 \approx 0.384 \)
• \( (0.103) \approx 0.103 \)

Putting it all together:

\[ P(9) \approx 10 \cdot 0.384 \cdot 0.103 \approx 0.3950 \]

Probability \( P(10) \)

\[ P(10) = \binom{10}{10} (0.897)^{10} (0.103)^{0} \]

Calculating the components:

• \( \binom{10}{10} = 1 \)
• \( (0.897)^{10} \approx 0.344\)
• \( (0.103)^{0} = 1 \)

Putting it all together:

\[ P(10) \approx 1 \cdot 0.344 \cdot 1 \approx 0.344 \]

Summary of Results

Finally, summarizing:

• \( P(6) \approx 0.01261 \)
• \( P(7) \approx 0.06307 \)
• \( P(8) \approx 0.20581 \)
• \( P(9) \approx 0.3950 \)
• \( P(10) \approx 0.344 \)

Note: Ensure the calculations were performed accurately, and check for rounding as needed throughout.