Resourcesdown chevron
Testimonials
Plans
Sign in
Sign up
Resourcesright arrow
Testimonials
Plans
Bytelearn - cat image with glassesAI tutor

Welcome to Bytelearn!

Let’s check out your problem:

Mr. and Mrs. Doran have a genetic history such that the probability that a child being born to them with a certain trait is 2//3. If they have five children, what is the probability that at least four of their five children will have that trait? Round your answer to the nearest thousandth. Answer:

Mr. and Mrs. Doran have a genetic history such that the probability that a child being born to them with a certain trait is 2/3 2 / 3 . If they have five children, what is the probability that at least four of their five children will have that trait? Round your answer to the nearest thousandth.\newlineAnswer:

View step-by-step helpView step-by-step help
Homeright chevron icon
Math Problemsright chevron icon
Algebra 2right chevron icon
Find probabilities using the binomial distributionright chevron icon

Full solution

Q. Mr. and Mrs. Doran have a genetic history such that the probability that a child being born to them with a certain trait is 2/3 2 / 3 . If they have five children, what is the probability that at least four of their five children will have that trait? Round your answer to the nearest thousandth.\newlineAnswer:
  1. Identify values for formula: Identify the values of nn, kk, and pp for the binomial probability formula.\newlinen=5n = 5 (the number of children)\newlinep=23p = \frac{2}{3} (the probability of a child having the trait)\newlineWe want to find the probability of at least four children having the trait, so we need to calculate the probability for k=4k = 4 and k=5k = 5.
  2. Use binomial probability formula: Use the binomial probability formula for k=4k = 4.P(X=k)=C(n,k)⋅(p)k⋅(1−p)(n−k)P(X = k) = C(n, k) \cdot (p)^k \cdot (1-p)^{(n-k)}Substitute n=5n = 5, k=4k = 4, and p=23p = \frac{2}{3} into the formula.P(X=4)=C(5,4)⋅(23)4⋅(1−23)(5−4)P(X = 4) = C(5, 4) \cdot \left(\frac{2}{3}\right)^4 \cdot \left(1 - \frac{2}{3}\right)^{(5 - 4)}
  3. Calculate binomial coefficient: Calculate the binomial coefficient C(5,4)C(5, 4).C(5,4)=5!4!(5−4)!C(5, 4) = \frac{5!}{4!(5 - 4)!}C(5,4)=51C(5, 4) = \frac{5}{1}C(5,4)=5C(5, 4) = 5
  4. Calculate (2/3)4(2/3)^4: Calculate (2/3)4(2/3)^4.
    (2/3)4=(2/3)×(2/3)×(2/3)×(2/3)(2/3)^4 = (2/3) \times (2/3) \times (2/3) \times (2/3)
    (2/3)4=16/81(2/3)^4 = 16/81
  5. Calculate (1−23)(5−4)(1 - \frac{2}{3})^{(5 - 4)}: Calculate (1−23)(5−4)(1 - \frac{2}{3})^{(5 - 4)}.(1−23)(5−4)=(13)1(1 - \frac{2}{3})^{(5 - 4)} = (\frac{1}{3})^1(1−23)(5−4)=13(1 - \frac{2}{3})^{(5 - 4)} = \frac{1}{3}
  6. Calculate P(X=4)P(X = 4): Calculate P(X=4)P(X = 4).P(X=4)=5×(1681)×(13)P(X = 4) = 5 \times \left(\frac{16}{81}\right) \times \left(\frac{1}{3}\right)P(X=4)=5×(1681)×(13)P(X = 4) = 5 \times \left(\frac{16}{81}\right) \times \left(\frac{1}{3}\right)P(X=4)=80243P(X = 4) = \frac{80}{243}
  7. Use binomial probability formula: Use the binomial probability formula for k=5k = 5.
    P(X=5)=C(5,5)⋅(23)5⋅(1−23)(5−5)P(X = 5) = C(5, 5) \cdot \left(\frac{2}{3}\right)^5 \cdot \left(1 - \frac{2}{3}\right)^{(5 - 5)}
    Substitute n=5n = 5, k=5k = 5, and p=23p = \frac{2}{3} into the formula.
    P(X=5)=C(5,5)⋅(23)5⋅(1−23)(0)P(X = 5) = C(5, 5) \cdot \left(\frac{2}{3}\right)^5 \cdot \left(1 - \frac{2}{3}\right)^{(0)}
  8. Calculate binomial coefficient: Calculate the binomial coefficient C(5,5)C(5, 5).C(5,5)=5!5!(5−5)!C(5, 5) = \frac{5!}{5!(5 - 5)!}C(5,5)=11C(5, 5) = \frac{1}{1}C(5,5)=1C(5, 5) = 1
  9. Calculate (2/3)5(2/3)^5: Calculate (2/3)5(2/3)^5.
    (2/3)5=(2/3)×(2/3)×(2/3)×(2/3)×(2/3)(2/3)^5 = (2/3) \times (2/3) \times (2/3) \times (2/3) \times (2/3)
    (2/3)5=32/243(2/3)^5 = 32/243
  10. Calculate (1−23)(5−5)(1 - \frac{2}{3})^{(5 - 5)}: Calculate (1−23)(5−5)(1 - \frac{2}{3})^{(5 - 5)}.\newline(1−23)(5−5)=(13)0(1 - \frac{2}{3})^{(5 - 5)} = (\frac{1}{3})^0\newline(1−23)(5−5)=1(1 - \frac{2}{3})^{(5 - 5)} = 1
  11. Calculate P(X=5)P(X = 5): Calculate P(X=5)P(X = 5).P(X=5)=1×(32243)×1P(X = 5) = 1 \times \left(\frac{32}{243}\right) \times 1P(X=5)=32243P(X = 5) = \frac{32}{243}
  12. Calculate probability of at least four children: Calculate the probability of at least four children having the trait.\newlineThis is the sum of the probabilities for k=4k = 4 and k=5k = 5.\newlineP(at least 4)=P(X=4)+P(X=5)P(\text{at least 4}) = P(X = 4) + P(X = 5)\newlineP(at least 4)=80243+32243P(\text{at least 4}) = \frac{80}{243} + \frac{32}{243}\newlineP(at least 4)=112243P(\text{at least 4}) = \frac{112}{243}
  13. Convert probability to decimal: Convert the probability to a decimal and round to the nearest thousandth.\newlineP(at least 4)=112243P(\text{at least } 4) = \frac{112}{243}\newlineP(at least 4)≈0.461P(\text{at least } 4) \approx 0.461

More problems from Find probabilities using the binomial distribution

Question
A restaurant server claims that \(28\)% of the time he leaves candy with the bill, the customer tips well. \newlineIf the server's claim is true, and one day he leaves candy with \(4\) customers' bills, what is the probability that exactly \(3\) of those customers will tip well?\newlineWrite your answer as a decimal rounded to the nearest thousandth.
Get tutor helpright-arrow

Posted 7 months ago

Question
There is a spinner with `50` equally likely sections, numbered from `1` to `50`. You have the opportunity to spin it. If the number is odd, you win $`11`. If the number is even, you win nothing. If you play the game, what is the expected payoff?\(\$\)_____
Get tutor helpright-arrow

Posted 8 months ago

Question
There are two different raffles you can enter.\newlineIn raffle A, one ticket will win a `$720` prize, and the other tickets will win nothing. There are 1,0001,000 in the raffle, each costing `$11`.\newlineRaffle B is for a `$370` prize. Out of 125125 tickets, each costing `$5`, one ticket will win the prize, and the other tickets will win nothing.\newlineWhich raffle is a better deal?\newlineChoices:\newline[A]Raffle A\text{[A]Raffle A}\newline[B]Raffle B\text{[B]Raffle B}
Get tutor helpright-arrow

Posted 8 months ago

Question
XX is a normally distributed random variable with mean 4949 and standard deviation 2525.\newlineWhat is the probability that XX is between 2424 and 7474??\newlineUse the 0.68−0.95−0.9970.68-0.95-0.997 rule and write your answer as a decimal. Round to the nearest thousandth if necessary.\newline____
Get tutor helpright-arrow

Posted 8 months ago

Question
XX is a normally distributed random variable with mean 6565 and standard deviation 88.\newlineWhat is the probability that XX is between 6262 and 6868??\newlineWrite your answer as a decimal rounded to the nearest thousandth.\newline____
Get tutor helpright-arrow

Posted 8 months ago

Question
Complete the statement. Round your answer to the nearest thousandth.\newlineIn a population that is normally distributed with mean 4343 and standard deviation 33, the bottom 30%30\% of the values are those less than ______.
Get tutor helpright-arrow

Posted 8 months ago

Question
Y Y is the number of desserts available in a randomly chosen restaurant.\newlineIs the random variable Y Y discrete or continuous?\newlineChoices:\newline[A]discrete \text{[A]discrete} \newline[B]continuous \text{[B]continuous}
Get tutor helpright-arrow

Posted 8 months ago

Question
Which of the following functions are continuous for all real numbers?\newlineg(x)=ln⁡(x) g(x)=\ln (x) \newlinef(x)=1x f(x)=\frac{1}{x} \newlineChoose 11 answer:\newline(A) g g only\newline(B) f f only\newline(C) Both g g and f f \newlineD Neither g g nor f f
Get tutor helpright-arrow

Posted 9 months ago

Question
Which of the following functions are continuous at x=−1 x=-1 ?\newlineg(x)=sin⁡(x+1) g(x)=\sin (x+1) \newlinef(x)=1x−1 f(x)=\frac{1}{x-1} \newlineChoose 11 answer:\newline(A) g g only\newline(B) f f only\newline(C) Both g g and f f \newline(D) Neither g g nor f f
Get tutor helpright-arrow

Posted 8 months ago

Question
Let g g be a continuous function on the closed interval [−1,4] [-1,4] , where g(−1)=−4 g(-1)=-4 and g(4)=1 g(4)=1 .\newlineWhich of the following is guaranteed by the Intermediate Value Theorem?\newlineChoose 11 answer:\newline(A) g(c)=−3 g(c)=-3 for at least one c c between −1-1 and 44\newline(B) g(c)=3 g(c)=3 for at least one c c between −4-4 and 11\newline(C) g(c)=−3 g(c)=-3 for at least one c c between −4-4 and 11\newline(D) g(c)=3 g(c)=3 for at least one c c between −1-1 and 44
Get tutor helpright-arrow

Posted 8 months ago

View all (44)

Related topics

Algebra - Order of OperationsAlgebra - Distributive Property`X` and `Y` AxesGeometry - Scalene TriangleCommon MultipleGeometry - Quadrant