The Binomial distributions is deal with discrete random variable - Free Printable
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Step-by-step solution for: The Binomial distributions is deal with discrete random variable
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Show Answer Key & Explanations
Step-by-step solution for: The Binomial distributions is deal with discrete random variable
Here are the step-by-step solutions for each problem on your assignment.
Step 1: Identify the probabilities.
* A standard die has numbers 1, 2, 3, 4, 5, 6.
* The even numbers are 2, 4, and 6. So, there are 3 even numbers out of 6 total.
* Probability of success (getting an even number), $p = \frac{3}{6} = 0.5$.
* Probability of failure (getting an odd number), $q = 1 - 0.5 = 0.5$.
Step 2: Identify the parameters.
* Number of trials ($n$) = 11.
* Number of successes needed ($k$) = 6.
Step 3: Use the Binomial Probability Formula.
$$P(X=k) = \binom{n}{k} p^k q^{n-k}$$
$$P(X=6) = \binom{11}{6} (0.5)^6 (0.5)^{11-6}$$
$$P(X=6) = \binom{11}{6} (0.5)^6 (0.5)^5$$
$$P(X=6) = \binom{11}{6} (0.5)^{11}$$
Step 4: Calculate.
* $\binom{11}{6} = \frac{11!}{6!(11-6)!} = \frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2 \times 1} = 462$
* $(0.5)^{11} \approx 0.00048828$
* $462 \times 0.00048828 \approx 0.2256$
Final Answer: 0.2256
---
Step 1: Identify the probabilities.
* $p = 0.5$ (making the shot)
* $q = 0.5$ (missing the shot)
Step 2: Identify the parameters.
* $n = 8$
* $k = 5$
Step 3: Use the formula.
$$P(X=5) = \binom{8}{5} (0.5)^5 (0.5)^{8-5}$$
$$P(X=5) = \binom{8}{5} (0.5)^8$$
Step 4: Calculate.
* $\binom{8}{5} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$
* $(0.5)^8 = 0.00390625$
* $56 \times 0.00390625 = 0.21875$
Final Answer: 0.21875
---
Step 1: Understand "at most".
"At most nine" means 0, 1, 2, ..., up to 9 girls. It is much easier to calculate the opposite (complement) and subtract from 1.
The opposite of "at most 9" is "10 or 11".
So, $P(\text{at most } 9) = 1 - [P(10) + P(11)]$.
Step 2: Identify parameters.
* $n = 11$
* $p = 0.5$ (girl)
* $q = 0.5$ (boy)
Step 3: Calculate P(10) and P(11).
* $P(10) = \binom{11}{10} (0.5)^{10} (0.5)^1 = 11 \times (0.5)^{11}$
* $P(11) = \binom{11}{11} (0.5)^{11} (0.5)^0 = 1 \times (0.5)^{11}$
Sum of $P(10) + P(11) = (11 + 1) \times (0.5)^{11} = 12 \times (0.5)^{11}$.
$(0.5)^{11} \approx 0.00048828$
$12 \times 0.00048828 \approx 0.005859$
Step 4: Subtract from 1.
$1 - 0.005859 = 0.994141$
Final Answer: 0.9941
---
Step 1: Understand "at least".
"At least two" means 2, 3, 4, ... up to 8.
It is easier to use the complement: $1 - [P(0) + P(1)]$.
("0 evens" or "1 even").
Step 2: Identify parameters.
* $n = 8$
* $p = 0.5$ (even number)
* $q = 0.5$ (odd number)
Step 3: Calculate P(0) and P(1).
* $P(0) = \binom{8}{0} (0.5)^0 (0.5)^8 = 1 \times 1 \times (0.5)^8 = 0.00390625$
* $P(1) = \binom{8}{1} (0.5)^1 (0.5)^7 = 8 \times (0.5)^8 = 8 \times 0.00390625 = 0.03125$
Sum $= 0.00390625 + 0.03125 = 0.03515625$
Step 4: Subtract from 1.
$1 - 0.03515625 = 0.96484375$
Final Answer: 0.9648
---
Note: This is a Hypergeometric distribution problem (sampling without replacement), not Binomial, because the total number of items is fixed and small.
Step 1: Identify totals.
* Total Math pages = 8
* Total English pages = 2
* Total pages = 10
* Pages recovered = 8
Step 2: Determine the condition.
He needs to recover all 8 Math pages. Since he recovers 8 pages total, this means he must pick exactly 8 Math pages and 0 English pages.
Step 3: Calculate combinations.
* Ways to choose 8 Math pages from 8: $\binom{8}{8} = 1$
* Ways to choose 0 English pages from 2: $\binom{2}{0} = 1$
* Total ways to choose any 8 pages from 10: $\binom{10}{8} = \frac{10 \times 9}{2 \times 1} = 45$
Step 4: Calculate probability.
$$P = \frac{\text{Ways to get desired outcome}}{\text{Total possible outcomes}}$$
$$P = \frac{1 \times 1}{45} = \frac{1}{45}$$
$$1 \div 45 \approx 0.0222$$
Final Answer: 0.0222 (or 1/45)
---
Note: This is also a Hypergeometric problem.
Step 1: Identify totals.
* Total wrenches in box = 11
* Wrenches needed (successes in population) = 5
* Wrenches grabbed (sample size) = 5
Step 2: Determine the condition.
He needs to grab exactly the 5 specific wrenches he needs.
Step 3: Calculate combinations.
* Ways to choose the 5 needed wrenches from the 5 available: $\binom{5}{5} = 1$
* Ways to choose the remaining 0 wrenches from the other 6 in the box: $\binom{6}{0} = 1$
* Total ways to choose any 5 wrenches from 11: $\binom{11}{5} = \frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2 \times 1} = 462$
Step 4: Calculate probability.
$$P = \frac{1 \times 1}{462} = \frac{1}{462}$$
$$1 \div 462 \approx 0.00216$$
Final Answer: 0.0022 (or 1/462)
---
Step 1: Identify totals.
* Girls = 7
* Boys = 8
* Total Students = 15
* Students picked = 8
Step 2: Determine the condition.
We want exactly 5 Boys. This implies we also pick 3 Girls (since $8 - 5 = 3$).
Step 3: Calculate combinations.
* Ways to pick 5 Boys from 8: $\binom{8}{5} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$
* Ways to pick 3 Girls from 7: $\binom{7}{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$
* Total ways to pick 8 students from 15: $\binom{15}{8} = 6435$
Step 4: Calculate probability.
$$P = \frac{56 \times 35}{6435}$$
$$P = \frac{1960}{6435}$$
$$1960 \div 6435 \approx 0.30458$$
Final Answer: 0.3046
---
Step 1: Identify totals.
* Cracked phones = 9
* Good phones = $13 - 9 = 4$
* Total phones = 13
* Phones sold (sample) = 6
Step 2: Determine the condition.
We want exactly 4 Cracked phones. This implies we also sell 2 Good phones ($6 - 4 = 2$).
Step 3: Calculate combinations.
* Ways to pick 4 Cracked from 9: $\binom{9}{4} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126$
* Ways to pick 2 Good from 4: $\binom{4}{2} = \frac{4 \times 3}{2 \times 1} = 6$
* Total ways to pick 6 phones from 13: $\binom{13}{6} = 1716$
Step 4: Calculate probability.
$$P = \frac{126 \times 6}{1716}$$
$$P = \frac{756}{1716}$$
$$756 \div 1716 \approx 0.44056$$
Final Answer: 0.4406
---
Step 1: Identify totals.
* Girls = 7
* Boys = 6
* Total Students = 13
* Students picked = 8
Step 2: Determine the condition.
We want exactly 4 Boys. This implies we also pick 4 Girls ($8 - 4 = 4$).
Step 3: Calculate combinations.
* Ways to pick 4 Boys from 6: $\binom{6}{4} = \frac{6 \times 5}{2 \times 1} = 15$
* Ways to pick 4 Girls from 7: $\binom{7}{4} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$
* Total ways to pick 8 students from 13: $\binom{13}{8} = 1287$
Step 4: Calculate probability.
$$P = \frac{15 \times 35}{1287}$$
$$P = \frac{525}{1287}$$
$$525 \div 1287 \approx 0.4079$$
Final Answer: 0.4079
---
Step 1: Identify totals.
* Black buttons = 9
* Brown buttons = 6
* Total buttons = 15
* Buttons picked = 7
Step 2: Determine the condition.
We want exactly 5 Black buttons. This implies we also pick 2 Brown buttons ($7 - 5 = 2$).
Step 3: Calculate combinations.
* Ways to pick 5 Black from 9: $\binom{9}{5} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126$
* Ways to pick 2 Brown from 6: $\binom{6}{2} = \frac{6 \times 5}{2 \times 1} = 15$
* Total ways to pick 7 buttons from 15: $\binom{15}{7} = 6435$
Step 4: Calculate probability.
$$P = \frac{126 \times 15}{6435}$$
$$P = \frac{1890}{6435}$$
$$1890 \div 6435 \approx 0.2937$$
Final Answer: 0.2937
1) A six-sided die is rolled eleven times. What is the probability that the die will show an even number exactly six times?
Step 1: Identify the probabilities.
* A standard die has numbers 1, 2, 3, 4, 5, 6.
* The even numbers are 2, 4, and 6. So, there are 3 even numbers out of 6 total.
* Probability of success (getting an even number), $p = \frac{3}{6} = 0.5$.
* Probability of failure (getting an odd number), $q = 1 - 0.5 = 0.5$.
Step 2: Identify the parameters.
* Number of trials ($n$) = 11.
* Number of successes needed ($k$) = 6.
Step 3: Use the Binomial Probability Formula.
$$P(X=k) = \binom{n}{k} p^k q^{n-k}$$
$$P(X=6) = \binom{11}{6} (0.5)^6 (0.5)^{11-6}$$
$$P(X=6) = \binom{11}{6} (0.5)^6 (0.5)^5$$
$$P(X=6) = \binom{11}{6} (0.5)^{11}$$
Step 4: Calculate.
* $\binom{11}{6} = \frac{11!}{6!(11-6)!} = \frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2 \times 1} = 462$
* $(0.5)^{11} \approx 0.00048828$
* $462 \times 0.00048828 \approx 0.2256$
Final Answer: 0.2256
---
2) A basketball player has a 50% chance of making each free throw. What is the probability that the player makes exactly five out of eight free throws?
Step 1: Identify the probabilities.
* $p = 0.5$ (making the shot)
* $q = 0.5$ (missing the shot)
Step 2: Identify the parameters.
* $n = 8$
* $k = 5$
Step 3: Use the formula.
$$P(X=5) = \binom{8}{5} (0.5)^5 (0.5)^{8-5}$$
$$P(X=5) = \binom{8}{5} (0.5)^8$$
Step 4: Calculate.
* $\binom{8}{5} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$
* $(0.5)^8 = 0.00390625$
* $56 \times 0.00390625 = 0.21875$
Final Answer: 0.21875
---
3) One day, eleven babies are born at a hospital. Assuming each baby has an equal chance of being a boy or girl, what is the probability that at most nine of the eleven babies are girls?
Step 1: Understand "at most".
"At most nine" means 0, 1, 2, ..., up to 9 girls. It is much easier to calculate the opposite (complement) and subtract from 1.
The opposite of "at most 9" is "10 or 11".
So, $P(\text{at most } 9) = 1 - [P(10) + P(11)]$.
Step 2: Identify parameters.
* $n = 11$
* $p = 0.5$ (girl)
* $q = 0.5$ (boy)
Step 3: Calculate P(10) and P(11).
* $P(10) = \binom{11}{10} (0.5)^{10} (0.5)^1 = 11 \times (0.5)^{11}$
* $P(11) = \binom{11}{11} (0.5)^{11} (0.5)^0 = 1 \times (0.5)^{11}$
Sum of $P(10) + P(11) = (11 + 1) \times (0.5)^{11} = 12 \times (0.5)^{11}$.
$(0.5)^{11} \approx 0.00048828$
$12 \times 0.00048828 \approx 0.005859$
Step 4: Subtract from 1.
$1 - 0.005859 = 0.994141$
Final Answer: 0.9941
---
4) A six-sided die is rolled eight times. What is the probability that the die will show an even number at least two times?
Step 1: Understand "at least".
"At least two" means 2, 3, 4, ... up to 8.
It is easier to use the complement: $1 - [P(0) + P(1)]$.
("0 evens" or "1 even").
Step 2: Identify parameters.
* $n = 8$
* $p = 0.5$ (even number)
* $q = 0.5$ (odd number)
Step 3: Calculate P(0) and P(1).
* $P(0) = \binom{8}{0} (0.5)^0 (0.5)^8 = 1 \times 1 \times (0.5)^8 = 0.00390625$
* $P(1) = \binom{8}{1} (0.5)^1 (0.5)^7 = 8 \times (0.5)^8 = 8 \times 0.00390625 = 0.03125$
Sum $= 0.00390625 + 0.03125 = 0.03515625$
Step 4: Subtract from 1.
$1 - 0.03515625 = 0.96484375$
Final Answer: 0.9648
---
5) Pranav is carrying eight pages of math homework and two pages of English homework. A gust of wind blows the pages out of his hands and he is only able to recover eight random pages. What is the probability that he recovers all of his math homework?
Note: This is a Hypergeometric distribution problem (sampling without replacement), not Binomial, because the total number of items is fixed and small.
Step 1: Identify totals.
* Total Math pages = 8
* Total English pages = 2
* Total pages = 10
* Pages recovered = 8
Step 2: Determine the condition.
He needs to recover all 8 Math pages. Since he recovers 8 pages total, this means he must pick exactly 8 Math pages and 0 English pages.
Step 3: Calculate combinations.
* Ways to choose 8 Math pages from 8: $\binom{8}{8} = 1$
* Ways to choose 0 English pages from 2: $\binom{2}{0} = 1$
* Total ways to choose any 8 pages from 10: $\binom{10}{8} = \frac{10 \times 9}{2 \times 1} = 45$
Step 4: Calculate probability.
$$P = \frac{\text{Ways to get desired outcome}}{\text{Total possible outcomes}}$$
$$P = \frac{1 \times 1}{45} = \frac{1}{45}$$
$$1 \div 45 \approx 0.0222$$
Final Answer: 0.0222 (or 1/45)
---
6) A mechanic working under a car requires five different size wrenches from his toolbox, which contains eleven different wrenches. Reaching for his toolbox, he grabs five of them at random. What is the probability that the mechanic has all of the wrenches he needs?
Note: This is also a Hypergeometric problem.
Step 1: Identify totals.
* Total wrenches in box = 11
* Wrenches needed (successes in population) = 5
* Wrenches grabbed (sample size) = 5
Step 2: Determine the condition.
He needs to grab exactly the 5 specific wrenches he needs.
Step 3: Calculate combinations.
* Ways to choose the 5 needed wrenches from the 5 available: $\binom{5}{5} = 1$
* Ways to choose the remaining 0 wrenches from the other 6 in the box: $\binom{6}{0} = 1$
* Total ways to choose any 5 wrenches from 11: $\binom{11}{5} = \frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2 \times 1} = 462$
Step 4: Calculate probability.
$$P = \frac{1 \times 1}{462} = \frac{1}{462}$$
$$1 \div 462 \approx 0.00216$$
Final Answer: 0.0022 (or 1/462)
---
7) A class has seven girls and eight boys. If the teacher randomly picks eight students, what is the probability that she will pick exactly five boys?
Step 1: Identify totals.
* Girls = 7
* Boys = 8
* Total Students = 15
* Students picked = 8
Step 2: Determine the condition.
We want exactly 5 Boys. This implies we also pick 3 Girls (since $8 - 5 = 3$).
Step 3: Calculate combinations.
* Ways to pick 5 Boys from 8: $\binom{8}{5} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$
* Ways to pick 3 Girls from 7: $\binom{7}{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$
* Total ways to pick 8 students from 15: $\binom{15}{8} = 6435$
Step 4: Calculate probability.
$$P = \frac{56 \times 35}{6435}$$
$$P = \frac{1960}{6435}$$
$$1960 \div 6435 \approx 0.30458$$
Final Answer: 0.3046
---
8) A shipment of thirteen smartphones contains nine with cracked screens. If sold in a random order, what is the probability that exactly four of the first six sold have cracked screens?
Step 1: Identify totals.
* Cracked phones = 9
* Good phones = $13 - 9 = 4$
* Total phones = 13
* Phones sold (sample) = 6
Step 2: Determine the condition.
We want exactly 4 Cracked phones. This implies we also sell 2 Good phones ($6 - 4 = 2$).
Step 3: Calculate combinations.
* Ways to pick 4 Cracked from 9: $\binom{9}{4} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126$
* Ways to pick 2 Good from 4: $\binom{4}{2} = \frac{4 \times 3}{2 \times 1} = 6$
* Total ways to pick 6 phones from 13: $\binom{13}{6} = 1716$
Step 4: Calculate probability.
$$P = \frac{126 \times 6}{1716}$$
$$P = \frac{756}{1716}$$
$$756 \div 1716 \approx 0.44056$$
Final Answer: 0.4406
---
9) A class has seven girls and six boys. If the teacher randomly picks eight students, what is the probability that she will pick exactly four boys?
Step 1: Identify totals.
* Girls = 7
* Boys = 6
* Total Students = 13
* Students picked = 8
Step 2: Determine the condition.
We want exactly 4 Boys. This implies we also pick 4 Girls ($8 - 4 = 4$).
Step 3: Calculate combinations.
* Ways to pick 4 Boys from 6: $\binom{6}{4} = \frac{6 \times 5}{2 \times 1} = 15$
* Ways to pick 4 Girls from 7: $\binom{7}{4} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$
* Total ways to pick 8 students from 13: $\binom{13}{8} = 1287$
Step 4: Calculate probability.
$$P = \frac{15 \times 35}{1287}$$
$$P = \frac{525}{1287}$$
$$525 \div 1287 \approx 0.4079$$
Final Answer: 0.4079
---
10) A jar contains nine black buttons and six brown buttons. If seven buttons are picked at random, what is the probability that exactly five of them are black?
Step 1: Identify totals.
* Black buttons = 9
* Brown buttons = 6
* Total buttons = 15
* Buttons picked = 7
Step 2: Determine the condition.
We want exactly 5 Black buttons. This implies we also pick 2 Brown buttons ($7 - 5 = 2$).
Step 3: Calculate combinations.
* Ways to pick 5 Black from 9: $\binom{9}{5} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126$
* Ways to pick 2 Brown from 6: $\binom{6}{2} = \frac{6 \times 5}{2 \times 1} = 15$
* Total ways to pick 7 buttons from 15: $\binom{15}{7} = 6435$
Step 4: Calculate probability.
$$P = \frac{126 \times 15}{6435}$$
$$P = \frac{1890}{6435}$$
$$1890 \div 6435 \approx 0.2937$$
Final Answer: 0.2937
Parent Tip: Review the logic above to help your child master the concept of binomial probability worksheet.
