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inna [77]
3 years ago
12

A jar contains 8 marbles numbered 1 through 8. an experiment consists of randomly selecting a marble from the jar, observing the

number drawn, and then randomly selecting a card from a standard deck and observing the suit of the card (hearts, diamonds, clubs, or spades). how many outcomes are in the sample space for this experiment? how many outcomes are in the event "an even number is drawn?" how many outcomes are in the event "a number more than 2 is drawn and a red card is drawn?" how many outcomes are in the event "a number less than 3 is drawn or a club is not drawn?"
Mathematics
1 answer:
zepelin [54]3 years ago
3 0
There are 8 possible outcomes for a marble being drawn and numbered. 
{1,2,3,4,5,6,7,8}
There are 4 possible outcomes for a card being selected from a standard deck.
{ <span>hearts, diamonds, clubs, spades}
So the number of outcomes in the sample space would be 8 x 4 = 32.

In the event "an even number is drawn", there are only 4 possible outcomes for a marble being drawn, {2,4,6,8}, whereas there are still 4 possible outcomes for a suit. So the number of outcomes in the event is 4 x 4 = 16.

</span><span>In the event "a number more than 2 is drawn and a red card is drawn", there are 6 possible outcomes for the marble being drawn, {3,4,5,6,7,8}, whereas there are only two possible suits for a card being selected as red, {heart, diamond}. So the number of outcomes in this event is 6 x 2 = 12.

In the event </span><span>"a number less than 3 is drawn or a club is not drawn", the number drawn could be 1 or 2 whereas a spade/heart/diamond could be selected. So the number of outcomes is 2 x 3 = 6.</span><span>

</span>
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Answer:

One fourth of the caps are blue with tassels.

Step-by-step explanation:

Take 7/11 of the caps with tassels, which boils down to (7/11)(11/28), or 1/4.

One fourth of the caps are blue with tassels.

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1/6

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8 0
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ABC is an equilateral triangle .Three distinct points are marked on each sides of the triangle. (not on the vertices). (i) How m
iVinArrow [24]

The number of lines and triangles illustrates combination

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  • 84 triangles can be drawn using the nine points

<h3>The number of lines that can be drawn</h3>

The given parameters are:

Total points, n = 9

Total points on each line, r = 2

The number of lines that can be drawn is then calculated  using the following combination formula

Lines = nCr

This gives

Lines = 9C2

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Lines = 36

Hence, 36 lines can be drawn to join the nine points

<h3>The number of triangles that can be drawn</h3>

The given parameters are:

Total points, n = 9

Total points on each triangle, r = 3

The number of triangles that can be drawn is then calculated  using the following combination formula

Triangles = nCr

This gives

Triangles = 9C3

Evaluate the combination expression

Triangles = 84

Hence, 84 triangles can be drawn using the nine points

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4 0
2 years ago
The Venn diagram shows the results of two events resulting from rolling a number cube.
gulaghasi [49]

Answer:   P(A\cap B)=\frac{1}{3}

P(A)=\frac{1}{3}

P(B)=1

P(A|B)=\frac{1}{3}

Step-by-step explanation:

From the given figure it can be seen that

Total number on cube= n(S)=6

Intersection of A and B = n(A ∩ B)= 2

Therefore,

P(A\cap B)=\frac{n(A\cap B)}{n(s)}=\frac{2}{6}=\frac{1}{3}

Also, The number of elements in A = 2

Therefore,

P(A)=\frac{n(A)}{n(s)}=\frac{2}{6}=\frac{1}{3}

Similarly, The number of elements in B= 6

Therefore,

P(B)=\frac{n(B)}{n(s)}=\frac{6}{6}=1

The formula to find the conditional probability is given by :-

P(A|B)=\frac{P(A\cap B)}{P(B)}\\\\\Rightarrow P(A|B)=\frac{\frac{1}{3}}{1}=\frac{1}{3}

4 0
4 years ago
Read 2 more answers
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