School violemee A survey found thot $$ 10 \% $$ of prodiems in schools were caused by volent actions of stindents. If a rantomly sefected sample of 10 sehoof problems were selected, find the following probabalies, Do not round micrmediate values. found the final anawer to three decimat places. Part $$ 0 / 3 $$ fint 1 at 1 (a) Exactly 5 were problems caused by volent actions. $$ P $$ (ekactly 5 were caused by volent actions) $$ = $$ $$ \square $$ 0.001 Part: $$ 1 / 3 $$ Part 2 of 3 (b) At least 3 were problems caused by violent actions. $$ P $$ (at least 3 were caused by violent actions) $$ = $$ $$ \square $$ $$ \square $$

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UpStudy AI · Accepted Answer

**Step 1. Define the parameters**

We are given a binomial experiment with  
- Number of trials: $$ n = 10 $$  
- Success probability: $$ p = 0.10 $$ (a problem is caused by violent actions)

In a binomial setting, the probability of exactly $$ k $$ successes is given by  
$$
P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}
$$

---

**Step 2. Part (a): Probability Exactly 5 are Caused by Violent Actions**

We need to find $$ P(X = 5) $$. Using the formula:
$$
P(X = 5) = \binom{10}{5} (0.10)^5 (0.90)^5
$$

1. Compute the combination:
   $$
   \binom{10}{5} = 252
   $$

2. Compute the powers:
   $$
   (0.10)^5 = 1 \times 10^{-5} \quad \text{and} \quad (0.90)^5 \approx 0.59049
   $$

3. Multiply the values:
   $$
   P(X = 5) \approx 252 \times (1 \times 10^{-5}) \times 0.59049 \approx 252 \times 0.0000059049 \approx 0.001487
   $$

Rounded to three decimal places:
$$
P(X = 5) \approx 0.001
$$

---

**Step 3. Part (b): Probability At Least 3 are Caused by Violent Actions**

We need to find $$ P(X \ge 3) $$. This is computed by:
$$
P(X \ge 3) = 1 - P(X \le 2)
$$
where
$$
P(X \le 2) = P(X = 0) + P(X = 1) + P(X = 2)
$$

1. Calculate $$ P(X = 0) $$:
   $$
   P(X = 0) = \binom{10}{0} (0.10)^0 (0.90)^{10} = (0.90)^{10} \approx 0.3486784401
   $$

2. Calculate $$ P(X = 1) $$:
   $$
   P(X = 1) = \binom{10}{1} (0.10)^1 (0.90)^9 = 10 \times 0.10 \times (0.90)^9
   $$
   Since $$ (0.90)^9 \approx 0.387420489 $$,
   $$
   P(X = 1) \approx 10 \times 0.10 \times 0.387420489 \approx 0.387420489
   $$

3. Calculate $$ P(X = 2) $$:
   $$
   P(X = 2) = \binom{10}{2} (0.10)^2 (0.90)^8 = 45 \times 0.01 \times (0.90)^8
   $$
   Since $$ (0.90)^8 \approx 0.43046721 $$,
   $$
   P(X = 2) \approx 45 \times 0.01 \times 0.43046721 \approx 0.1937102445
   $$

4. Sum the probabilities for $$ X \le 2 $$:
   $$
   P(X \le 2) \approx 0.3486784401 + 0.387420489 + 0.1937102445 \approx 0.9298091736
   $$

5. Now find $$ P(X \ge 3) $$:
   $$
   P(X \ge 3) = 1 - 0.9298091736 \approx 0.0701908264
   $$

Rounded to three decimal places:
$$
P(X \ge 3) \approx 0.070
$$

---

**Final Answers:**

- (a) $$ P(\text{exactly 5}) \approx 0.001 $$
- (b) $$ P(\text{at least 3}) \approx 0.070 $$
$$ P(\text{exactly 5}) \approx 0.001 $$ $$ P(\text{at least 3}) \approx 0.070 $$

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