SOLUTION: Good day. please help me solve this problems.
Consider a bag containing 5 blue marbles and 3 green marbles.
I. Construct a tree diagram to represent all possible outcomes of dr
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-> SOLUTION: Good day. please help me solve this problems.
Consider a bag containing 5 blue marbles and 3 green marbles.
I. Construct a tree diagram to represent all possible outcomes of dr
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Question 1208266: Good day. please help me solve this problems.
Consider a bag containing 5 blue marbles and 3 green marbles.
I. Construct a tree diagram to represent all possible outcomes of drawing two marbles when the marble is replaced after each draw. [4 marks]
II. What is the probability of drawing two blue marbles in a row? [2 marks]
III. What is the probability of selecting a colour of each? [2 marks]
IV. Construct a tree diagram to represent all possible outcomes of drawing two marbles without replacement. [4 marks]
This is one way to draw the probability tree diagram.
This style has it flow from left to right.
Another diagram style would have it flow from top to bottom (rotate the diagram shown above 90 degrees clockwise).
The probability tree should be self-explanatory.
Perhaps the only explanation that's needed is each item in the "outcome" column is the result of multiplying the fractions along the particular pathway.
For instance, multiply the fractions along the upper most path to get (5/8)*(5/8) = 25/64 which is the probability of selecting 2 blue marbles in a row where replacement is done.
The list of possible outcomes are
P(BB) = 25/64
P(BG) = 15/64
P(GB) = 15/64
P(GG) = 9/64
Optionally we can make a table of outcomes
2nd is B
2nd is G
1st is B
25/64
15/64
1st is G
15/64
9/64
Note that P(BB)+P(BG)+P(GB)+P(GG) = (25/64)+(15/64)+(15/64)+(9/64) = 1
There are two methods we could use.
The first method is most direct and quickest.
P(one of each color) = P(BG) + P(GB)
P(one of each color) = (15/64)+(15/64)
P(one of each color) = (15+15)/64
P(one of each color) = 30/64
P(one of each color) = (2*15)/(2*32)
P(one of each color) = 15/32
Here is another approach.
P(two of same color) = P(2 blue OR 2 green)
P(two of same color) = P(BB)+P(GG)
P(two of same color) = (25/64)+(9/64)
P(two of same color) = (25+9)/64
P(two of same color) = 34/64
P(two of same color) = (2*17)/(2*32)
P(two of same color) = 17/32
Then,
P(one of each color) = 1 - P(two of same color)
P(one of each color) = 1 - (17/32)
P(one of each color) = (32/32) - (17/32)
P(one of each color) = (32-17)/32
P(one of each color) = 15/32
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Part IV
This probability tree is very similar to the tree in part I.
The 5/8 and 3/8 in the 1st column, or left-most column, are the same as before.
The fractions in the 2nd column are new. The 7s in the denominator refer to the number of marbles remaining after the 1st selection is not replaced.
Subtract 1 from the numerator only when you are selecting two of the same color.
To get each item in the "outcome" column, you follow the same idea as mentioned earlier.
Example: Multiplying along the top most pathway yields (5/8)*(4/7) = 20/56 = 5/14 which is the probability of getting two blue marbles in a row when replacement is not applied.
Use this idea for the other three pathways.
Here is an optional table of outcomes
2nd is B
2nd is G
1st is B
5/14
15/56
1st is G
15/56
3/28
Or we can list them out like so
P(BB) = 5/14
P(BG) = 15/56
P(GB) = 15/56
P(GG) = 3/28
Note that P(BB)+P(BG)+P(GB)+P(GG) = (5/14)+(15/56)+(15/56)+(3/28) = 1
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Summary of answers:
Diagram is shown above.
probability of two blue in a row = 25/64 (with replacement)
probability one of each color = 15/32 (with replacement)
(