To calculate the probability that Mike will make exactly two out of the three additional paper wad shots, we use the binomial probability formula:

$P(X = k) = \binom{n}{k} \cdot p^k \cdot (1-p)^{n-k}$

Where:

• $n$ = total number of trials (in this case, 3),
• $k$ = number of successes (in this case, 2),
• $p$ = probability of success on a single trial (60% = 0.6),
• $\binom{n}{k}$ = the number of ways to choose $k$ successes from $n$ trials, calculated as:

$\binom{n}{k} = \frac{n!}{k!(n-k)!}$

Step-by-Step Solution:

1. Calculate $\binom{n}{k}$: $\binom{3}{2} = \frac{3!}{2!(3-2)!} = \frac{3 \cdot 2 \cdot 1}{2 \cdot 1 \cdot 1} = 3$

2. Substitute into the binomial formula: $P(X = 2) = \binom{3}{2} \cdot p^2 \cdot (1-p)^{3-2}$ Substituting $p = 0.6$: $P(X = 2) = 3 \cdot (0.6)^2 \cdot (1-0.6)^1$

3. Simplify the terms:

   • $(0.6)^2 = 0.36$,
   • $1-0.6 = 0.4$,
   • $0.4^1 = 0.4$.

   So: $P(X = 2) = 3 \cdot 0.36 \cdot 0.4$

4. Calculate the result: $P(X = 2) = 3 \cdot 0.144 = 0.432$

Final Answer:

The probability that Mike will make exactly 2 out of the 3 additional shots is 0.432, or 43.2%.

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Would you like a detailed explanation of the binomial formula or its applications? 😊

Related Questions:

1. What is the probability Mike will make all 3 shots?
2. How does the probability change if Mike’s success rate improves to 70%?
3. What is the expected number of successful shots out of 3?
4. How would the probability distribution for 3 shots look graphically?
5. What is the variance and standard deviation for Mike’s performance over 3 shots?

Tip: The binomial formula is powerful for solving discrete probability problems involving repeated independent trials!