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2 years ago

7

Tow ordinary fair dice are rolled completely in the tree diagram. Work out the probability that exactly one of the dice lands on

a number less than 3? The last question (b) is what I need help with. 20 POINTS!

1 answer:

Lostsunrise [7]2 years ago

6 0

Answer:

1/3 (one third)

Because there are only 4 numbers on 2 die that have numbers less than 3 so 4 twelfths equals one third

Step-by-step explanation:

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Is 2/12 a terminating term? ?

boyakko [2]

Hi pupil here's your answer ::

____________________________

This term 2/11 is not a terminating one because when we divide 2 by 11 the answer is 0.1818181818181818181818181818181818181818...........and so on. So the answer is non obtaining that means that it is an non terminating term.

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hope that it helps. . . . . .

3 0

3 years ago

Please simplify picture attached

Mekhanik [1.2K]

Answer:

It is the first bubble.

Step-by-step explanation:

multiply the fraction $\frac{\sqrt{5} }{\sqrt{} 5}$

multiply the numerators and denominators seperatly

calculate the product

multiply the numbers

solution

3 0

3 years ago

Read 2 more answers

Find the equation of the line that<br> is parallel to y = 5x – 4 and<br> contains the point (4,5).

lilavasa [31]

Answer:

y = 3+1/2

that should be the answer

6 0

3 years ago

Tayâ€“Sachs Disease Tayâ€“Sachs disease is a genetic disorder that is usually fatal in young children. If both parents are carri

BARSIC [14]

Answer:

a) 0.0156 = 1.56% probability that all children will develop the disease.

b) 0.4219 = 42.19% probability that only one child will develop the disease.

c) 0.1406 = 14.06% probability that the third children will develop the disease, given that the first two did not.

Step-by-step explanation:

For each children, there are only two possible outcomes. Either they carry the disease, or they do not. The probability of a children carrying the disease is independent of any other children, which means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

The probability that their offspring will develop the disease is approximately .25.

This means that $p = 0.25$

Three children:

This means that $n = 3$

Question a:

This is P(X = 3). So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 3) = C_{3,3}.(0.25)^{3}.(0.75)^{0} = 0.0156$

0.0156 = 1.56% probability that all children will develop the disease.

Question b:

This is P(X = 1). So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 1) = C_{3,1}.(0.25)^{1}.(0.75)^{2} = 0.4219$

0.4219 = 42.19% probability that only one child will develop the disease.

c. The third child will develop Tayâ€“Sachs disease, given that the first two did not.

Third independent of the first two, so just multiply the probabilities.

First two do not develop, each with 0.75 probability.

Third develops, which 0.25 probability. So

$p = 0.75*0.75*0.25 = 0.1406$

0.1406 = 14.06% probability that the third children will develop the disease, given that the first two did not.

6 0

3 years ago

Students in a statistics class are conducting a survey to estimate the mean number of units students at their college are enroll

Papessa [141]

Answer:

The 95% confidence interval for the average number of units that students in their college are enrolled in is between 11.7 and 12.5 units.

Step-by-step explanation:

We have the standard deviation for the sample, so we use the t-distribution to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 45 - 1 = 44

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 44 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.95}{2} = 0.975$ . So we have T = 2

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 2\frac{1.5}{\sqrt{45}} = 0.4$

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 12.1 - 0.4 = 11.7 units

The upper end of the interval is the sample mean added to M. So it is 12.1 + 0.4 = 12.5 units

The 95% confidence interval for the average number of units that students in their college are enrolled in is between 11.7 and 12.5 units.

8 0

3 years ago