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3 years ago

In a family with 5 children, excluding multiple births, what is the probability of having exactly 2 girls?

Mathematics

2 answers:

svetoff [14.1K]3 years ago

6 0

1/2*1/2*1/2*1/2*1/2=1/32

Nat2105 [25]3 years ago

4 0

There is a 1/32 chance of that family having a girl.

You might be interested in

if you roll a fair 6-sided die 9 times, what is the probability that at least 2 of the rolls come up as a 3 or a 4?

Kay [80]

Using the binomial distribution, it is found that there is a 0.857 = 85.7% probability that at least 2 of the rolls come up as a 3 or a 4.

For each die, there are only two possible outcomes, either a 3 or a 4 is rolled, or it is not. The result of a roll is independent of any other roll, hence, the <em>binomial distribution</em> is used to solve this question.

Binomial probability distribution

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

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

The parameters are:

x is the number of successes.

n is the number of trials.

p is the probability of a success on a single trial.

In this problem:

There are 9 rolls, hence $n = 9$ .

Of the six sides, 2 are 3 or 4, hence $p = \frac{2}{6} = 0.3333$

The desired probability is:

$P(X \geq 2) = 1 - P(X < 2)$

In which:

$P(X < 2) = P(X = 0) + P(X = 1)$

Then

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

$P(X = 0) = C_{9,0}.(0.3333)^{0}.(0.6667)^{9} = 0.026$

$P(X = 1) = C_{9,1}.(0.3333)^{1}.(0.6667)^{8} = 0.117$

Then:

$P(X < 2) = P(X = 0) + P(X = 1) = 0.026 + 0.117 = 0.143$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.143 = 0.857$

0.857 = 85.7% probability that at least 2 of the rolls come up as a 3 or a 4.

For more on the binomial distribution, you can check brainly.com/question/24863377

7 0

3 years ago

What is the slope of a line that passes through the points (5,0) and (10,0)?

disa [49]

Answer:

Step-by-step explanation:

Use two-point form to find the slope.

m=y2-y1/x2-x1

m=0-0/10-5

m=0/5

4 0

3 years ago

in chemistry class Andrew has earned scores of 64 69 and 73 on three tests he must maintain an average of 72

olganol [36]

You haven't told me what the question is. But I put the mouse
to my forehead, closed my eyes, took a deep breath, and I could
see it shimmering in my mind's eye. It was quite fuzzy, but I think
the question is

   "What score does Andrew need on the next test
   in order to raise his average to 72% ?"

The whole experience drew an incredible amount of energy
out of me, and the mouse is a total wreck. So we'll just go ahead
and answer that one. I hope it's the correct question.

The average score on 4 tests is

   (1/4) (the sum of all the scores) .

In order for Andrew to have a 72% average on 4 tests,
the sum of the 4 scores must be

   (4) x (72%) = 288% .

Out of that total that he needs, he already has

(64% + 69% + 73%) = 206%

on the first three tests.

So in order to average 72% for all 4 tests,
he'll need to score

   (288% - 206%) = 82%

on the fourth one.

6 0

3 years ago

Triangle A″B″C″ is formed by a reflection over x = −3 and dilation by a scale factor of 3 from the origin. Which equation shows

Allisa [31]

Answer:

see below

Step-by-step explanation:

Each segment in ΔA"B"C" is 3 times the length of the corresponding segment in ΔABC. This is due to the dilation by a scale factor of 3.

Then you have ...

$\overline{A''B''}=3\overline{AB}\quad\text{or}\\\\\dfrac{\overline{AB}}{\overline{A''B''}}=\dfrac{1}{3}$