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1 year ago

Randomly meeting a four child family with either one or exactly 2 boy children

Mathematics

1 answer:

Lelechka [254]1 year ago

3 0

Answer:

The probability of randomly meeting a four child family with either exactly one or exactly two boy children = (5/8) = 0.625

Step-by-step explanation:

Complete Question

The probability of randomly meeting a four child family with either exactly one or exactly two boy children.

Solution

The possible sample spaces for a family with four children include

4 boys and 0 Girl

BBBB

3 boys and 1 girl

BBBG BBGB BGBB GBBB

2 boys and 2 girls

BBGG BGBG BGGB GBBG GBGB GGBB

1 boy and 3 girls

BGGG GBGG GGBG GGGB

0 boy and 4 girls

GGGG

Total number of elements in the sample space = 16

Probability of an event is defined as the number of elements in that event divided by the Total number of elements in the sample space.

The required probability is a sum of probabilities.

The probability of meeting a four child family with exactly 1 boy = (4/16) = (1/4) = 0.25

The probability of meeting a four child family with exactly 2 boys = (6/16) = (3/8) = 0.375

The probability of randomly meeting a four child family with either exactly one or exactly two boy children = (4/16) + (6/16) = (10/16) = (5/8) = 0.25 + 0.375 = 0.625

Hope this Helps!!!

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1 year ago

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The speed limit on a road is 60 mph. The actual speeds of cars are measured and found to be normally distributed with a mean val

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Answer: 8.1%

Step-by-step explanation:

Given : $\mu=63$

$\sigma=5$

Let x be the random variable that represents the actual speeds of cars.

The speed limit on a road is 60 mph.

Using formula , $z=\dfrac{x-\mu}{\sigma}$ , we have for x= 60+10=70

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2 years ago

For a polygon to be convex means that all of its interior angles are less than 180 degrees. Prove that for all integers n ≥ 3, t

Alexeev081 [22]

Answer: check explanation

Step-by-step explanation:

Let K(n) be the sum of the interior angles in any n-sided convex polygon which is exactly 180(n −2)

degrees.

CASE: n = 3. A 3-sided polygon is a triangle, whose interior angles were shown always to sum to be 180 degree

INDUCTION HYPOTHESIS: Suppose that K(b) holds for some b ≥ 3. Which means that the interior angles in any b-sided convex

polygon is exactly 180(n −2) degrees.

INDUCTION STEP: We need to show that K(n is greater than or equals to 3) . That is, the interior angles of any b+1-sided convex polygon

is exactly 180(b−2) = 180(b −1) degrees,

Let X be any (b+1)-vertex convex polygon, say with successive vertices x1, x2,..., xb+

Now Y is also a convex polygon , so by the induction hypothesis K(b), the sum of the interior

angles of Y is 180(k −2).

Now let T be the triangle with vertices xk , xb+1, x1. The sum of the interior angles in X is the sum of those

in Y plus the sum of those in T .

So the sum of the interior angles in X is

180(b −2)+180 = 180((b +1)−2) = 180(b −1).

Since X was arbitrary, we conclude that the sum of the interior angles of any (b +1)-sided convex polygon

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1 year ago

Consider the midterm and final for a statistics class. Suppose 13% of students earned an A on the midterm. Of those students who

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Answer:

There is a 38.97% probability that this student earned an A on the midterm.

Step-by-step explanation:

The first step is that we have to find the percentage of students who got an A on the final exam.

Suppose 13% students earned an A on the midterm. Of those students who earned an A on the midterm, 47% received an A on the final, and 11% of the students who earned lower than an A on the midterm received an A on the final.

This means that

Of the 13% of students who earned an A on the midterm, 47% received an A on the final. Also, of the 87% who did not earn an A on the midterm, 11% received an A on the final.

So, the percentage of students who got an A on the final exam is

$P_{A} = 0.13(0.47) + 0.87(0.11) = 0.1568$

To find the probability that this student earned an A on the final test also earned on the midterm, we divide the percentage of students who got an A on both tests by the percentage of students who got an A on the final test.

The percentage of students who got an A on both tests is:

$P_{AA} = 0.13(0.47) = 0.0611$