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Mariulka [41]
3 years ago
6

Maria has eight black marbles, fourteen clear marbles, and twelve blue marbles in a bag. If she picks two marbles at random, wit

hout replacement, what is the probability that she will select a blue marble first, then a clear marble?
Mathematics
1 answer:
horrorfan [7]3 years ago
8 0

Answer:

\boxed{0.15}

Step-by-step explanation:

Part 1: Solve for the total amount of marbles

To solve for the probability of certain events, a population is needed to derive this information from. In order to find this population, add up the amounts of each marble.

8 + 14 + 12 = 34 marbles

Part 2: Determine the probabilities

Now, given the amounts of marbles, simply multiply the ratios of blue marbles to total marbles and the ratio of clear marbles to total marbles to get the combined probability.

\frac{12}{34}*\frac{14}{33} =  \frac{28}{187} \approxeq 0.1497 \approxeq 0.15 * 100 = 15

The probability of these events occurring simultaneously is 15%.

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Write the equation of a line through (0,2) that is parallel to the graph
Vinvika [58]

Answer:

y=3x+2

Step-by-step explanation:

Refer to the picture that given :)

7 0
3 years ago
Items are inspected for flaws by two quality inspectors. Both inspectors inspect every item and the probability that an item has
Genrish500 [490]

Answer:

a)0.976

b)0.00926

c)0.2402

d)0.35

Step-by-step explanation:

Let X_i be an item passed by inspector i

Let Y be the event that there is a fault in an item

The probability that an item has a flaw is 0.1 i.e. P(Y)=0.1

If a flaw is present ,it will be detected by the first inspector with probability 0.92 i.e.P(\bar{X_1}|Y)=0.92

So, P(X_1|Y)=1-0.92=0.08

If a flaw is present ,it will be detected by the second inspector with probability 0.7 i.e.P(\bar{X_2}|Y)=0.7

So,P(X_2|Y)=1-0.7=0.3

If an item does not have a flaw, it will be passed by the first inspector with probability 0.95 i.e. P(X_1|\bar{Y}) = 0.95

So, P(\bar{X_1}|\bar{Y}) = 1-0.95=0.05

If an item does not have a flaw, it will be passed by the second inspector with probability 0.8 i.e. P(X_2|\bar{Y}) = 0.8

So, P(\bar{X_2}|\bar{Y}) = 1-0.8=0.2

a)P(found by atleast one inspector | It has flaw )=1-P(found by none inspector | It has flow )

P(found by atleast one inspector | It has flaw )=1-P(X_1|Y) P(X_2|Y)

P(found by atleast one inspector | It has flaw )=1-0.08 \times 0.3

P(found by atleast one inspector | It has flaw )=0.976

Hence the probability that it will be found by at least one of the two inspectors if it has flaw is 0.976

b)P(Y|X_1)=\frac{P(X_1|Y) P(Y)}{P(X_1|Y) P(Y)+P(X_1|\bar{Y}) P(\bar{Y})}

P(Y|X_1)=\frac{0.08 \times 0.1}{0.08 \times 0.1+0.95 \times 0.9}=0.00926

C)P( two inspectors draw different conclusions on the same item)=P(X_1 \cap \bar{X_2} \cap Y)+P(\bar{X_1} \cap X_2 \cap Y)+P(X_1 \cap \bar{X_2} \cap \bar{Y})+P(\bar{X_1} \cap X_2 \cap \bar{Y})

P( two inspectors draw different conclusions on the same item)=0.2402

D)

P(Y|(X_1 \cap X_2))=\frac{P(Y \cap X_1 \cap X_2)}{P(X_1 \cap X_2)}\\P(Y|(X_1 \cap X_2))=\frac{P(Y \cap X_1 \cap X_2)}{P(X_1 \cap X_2 \cap Y)+P(X_1 \cap X_2 \cap \bar{Y})}\\P(Y|(X_1 \cap X_2))=0.35

3 0
3 years ago
Evaluate the function for x = –2a, if a = 3.<br> f(x)= -4x+7
wariber [46]
I got -17; hoped you liked it
7 0
3 years ago
Angle XYZ is bisected by line YW the measure of angle XYW is 52 what is the measurement of angle WYZ what is the measurement of
LekaFEV [45]
Bisected- split into two equal parts
XYW=WYZ
XYW-52°
answer= WYZ-52°
52+52= 104
answer= XYZ-104°

5 0
3 years ago
Using pemdas what’s the correct solution
SSSSS [86.1K]

4- 2/3 (4-1/6) divided by 3/4

parenthesis first

4 - 2/3 (3 5/6) divided by 3/4

change to an improper fraction (6*3+5)/6

4 - 2/3 ( 23/6)divided by 3/4

4 - 46/18 divide by3/4

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4 - 46/18 * 4/3

4 - 23/9 * 4/3

4 - 92/27

get a common denominator of 27

4*27/27 -92/27

108/27 - 92/27

16/27

4 0
4 years ago
Read 2 more answers
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