Given that X is a binomial distribution with n=8 and p=0.6. The probability mass function is P(X=k) = nk p^k (1-p)^n-k.

Question

Given that X is a binomial distribution with n=8 and p=0.6. The probability mass function is P(X=k) = nk p^k (1-p)^n-k.

Asked on March 27, 2026Computer Science

This computer science problem involves algorithmic thinking and programming concepts. The solution below explains the approach, logic, and implementation step by step.

Accepted Answer · 2026-03-27T02:13:09.167Z

Given that $X$ is a binomial distribution with $n=8$ and $p=0.6$. The probability mass function is $P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$. Here, $n=8$, $p=0.6$, and $1-p = 0.4$. a) Find $P(X=5)$ Step 1: Apply the binomial probability formula. $$P(X=5) = \binom{8}{5} (0.6)^5 (0.4)^{8-5}$$ $$P(X=5) = \binom{8}{5} (0.6)^5 (0.4)^3$$ Step 2: Calculate the binomial coefficient. $$\binom{8}{5} = \frac{8!}{5!(8-5)!} = \frac{8!}{5!3!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$$ Step 3: Calculate the powers of $p$ and $(1-p)$. $$(0.6)^5 = 0.07776$$ $$(0.4)^3 = 0.064$$ Step 4: Multiply the values. $$P(X=5) = 56 \times 0.07776 \times 0.064 = 0.27869184$$ Step 5: Round to 4 decimal places. $$P(X=5) \approx 0.2787$$ The final answer is $\boxed{\text{0.2787}}$ b) Find $P(X > 6)$ Step 1: Express $P(X > 6)$ as a sum of probabilities. $$P(X > 6) = P(X=7) + P(X=8)$$ Step 2: Calculate $P(X=7)$. $$P(X=7) = \binom{8}{7} (0.6)^7 (0.4)^{8-7} = \binom{8}{7} (0.6)^7 (0.4)^1$$ $$\binom{8}{7} = 8$$ $$(0.6)^7 = 0.0279936$$ $$P(X=7) = 8 \times 0.0279936 \times 0.4 = 0.08957952$$ Step 3: Calculate $P(X=8)$. $$P(X=8) = \binom{8}{8} (0.6)^8 (0.4)^{8-8} = \binom{8}{8} (0.6)^8 (0.4)^0$$ $$\binom{8}{8} = 1$$ $$(0.6)^8 = 0.01679616$$ $$P(X=8) = 1 \times 0.01679616 \times 1 = 0.01679616$$ Step 4: Sum the probabilities. $$P(X > 6) = 0.08957952 + 0.01679616 = 0.10637568$$ Step 5: Round to 4 decimal places. $$P(X > 6) \approx 0.1064$$ The final answer is $\boxed{\text{0.1064}}$ c) Find $P(2 < X < 6)$ Step 1: Express $P(2 < X < 6)$ as a sum of probabilities. $$P(2 < X < 6) = P(X=3) + P(X=4) + P(X=5)$$ Step 2: Calculate $P(X=3)$. $$P(X=3) = \binom{8}{3} (0.6)^3 (0.4)^{8-3} = \binom{8}{3} (0.6)^3 (0.4)^5$$ $$\binom{8}{3} = \frac{8!}{3!5!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$$ $$(0.6)^3 = 0.216$$ $$(0.4)^5 = 0.01024$$ $$P(X=3) = 56 \times 0.216 \times 0.01024 = 0.12386304$$ Step 3: Calculate $P(X=4)$. $$P(X=4) = \binom{8}{4} (0.6)^4 (0.4)^{8-4} = \binom{8}{4} (0.6)^4 (0.4)^4$$ $$\binom{8}{4} = \frac{8!}{4!4!} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70$$ $$(0.6)^4 = 0.1296$$ $$(0.4)^4 = 0.0256$$ $$P(X=4) = 70 \times 0.1296 \times 0.0256 = 0.2322432$$ Step 4: Use $P(X=5)$ from part a). $$P(X=5) = 0.27869184$$ Step 5: Sum the probabilities. $$P(2 < X < 6) = 0.12386304 + 0.2322432 + 0.27869184 = 0.63479808$$ Step 6: Round to 4 decimal places. $$P(2 < X < 6) \approx 0.6348$$ The final answer is $\boxed{\text{0.6348}}$

Accepted Answer · 2026-03-27T02:13:09.167Z

Given that $X$ is a binomial distribution with $n=8$ and $p=0.6$. The probability mass function is $P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$. Here, $n=8$, $p=0.6$, and $1-p = 0.4$. a) Find $P(X=5)$ Step 1: Apply the binomial probability formula. $$P(X=5) = \binom{8}{5} (0.6)^5 (0.4)^{8-5}$$ $$P(X=5) = \binom{8}{5} (0.6)^5 (0.4)^3$$ Step 2: Calculate the binomial coefficient. $$\binom{8}{5} = \frac{8!}{5!(8-5)!} = \frac{8!}{5!3!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$$ Step 3: Calculate the powers of $p$ and $(1-p)$. $$(0.6)^5 = 0.07776$$ $$(0.4)^3 = 0.064$$ Step 4: Multiply the values. $$P(X=5) = 56 \times 0.07776 \times 0.064 = 0.27869184$$ Step 5: Round to 4 decimal places. $$P(X=5) \approx 0.2787$$ The final answer is $\boxed{\text{0.2787}}$ b) Find $P(X > 6)$ Step 1: Express $P(X > 6)$ as a sum of probabilities. $$P(X > 6) = P(X=7) + P(X=8)$$ Step 2: Calculate $P(X=7)$. $$P(X=7) = \binom{8}{7} (0.6)^7 (0.4)^{8-7} = \binom{8}{7} (0.6)^7 (0.4)^1$$ $$\binom{8}{7} = 8$$ $$(0.6)^7 = 0.0279936$$ $$P(X=7) = 8 \times 0.0279936 \times 0.4 = 0.08957952$$ Step 3: Calculate $P(X=8)$. $$P(X=8) = \binom{8}{8} (0.6)^8 (0.4)^{8-8} = \binom{8}{8} (0.6)^8 (0.4)^0$$ $$\binom{8}{8} = 1$$ $$(0.6)^8 = 0.01679616$$ $$P(X=8) = 1 \times 0.01679616 \times 1 = 0.01679616$$ Step 4: Sum the probabilities. $$P(X > 6) = 0.08957952 + 0.01679616 = 0.10637568$$ Step 5: Round to 4 decimal places. $$P(X > 6) \approx 0.1064$$ The final answer is $\boxed{\text{0.1064}}$ c) Find $P(2 < X < 6)$ Step 1: Express $P(2 < X < 6)$ as a sum of probabilities. $$P(2 < X < 6) = P(X=3) + P(X=4) + P(X=5)$$ Step 2: Calculate $P(X=3)$. $$P(X=3) = \binom{8}{3} (0.6)^3 (0.4)^{8-3} = \binom{8}{3} (0.6)^3 (0.4)^5$$ $$\binom{8}{3} = \frac{8!}{3!5!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$$ $$(0.6)^3 = 0.216$$ $$(0.4)^5 = 0.01024$$ $$P(X=3) = 56 \times 0.216 \times 0.01024 = 0.12386304$$ Step 3: Calculate $P(X=4)$. $$P(X=4) = \binom{8}{4} (0.6)^4 (0.4)^{8-4} = \binom{8}{4} (0.6)^4 (0.4)^4$$ $$\binom{8}{4} = \frac{8!}{4!4!} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70$$ $$(0.6)^4 = 0.1296$$ $$(0.4)^4 = 0.0256$$ $$P(X=4) = 70 \times 0.1296 \times 0.0256 = 0.2322432$$ Step 4: Use $P(X=5)$ from part a). $$P(X=5) = 0.27869184$$ Step 5: Sum the probabilities. $$P(2 < X < 6) = 0.12386304 + 0.2322432 + 0.27869184 = 0.63479808$$ Step 6: Round to 4 decimal places. $$P(2 < X < 6) \approx 0.6348$$ The final answer is $\boxed{\text{0.6348}}$