Three friends play marbles each week. When they combine their marbles,

Question

Three friends play marbles each week. When they combine their marbles, they have 100 in total. 45 of the marbles are new and the rest are old. 30 are red, 20 are green, 25 are yellow, and the rest are white. What is the probability that a randomly chosen marble is new OR yellow?

solution:-

P (New or Yellow) = P(New) + P(Yellow) - P(New and Yellow)
= 45/100 + 25/100 - (45/100 * 25/100)
= 47/80

option E

A.  \frac{4}{53}

B.  \frac{3}{13}

C.  \frac{7}{15}

D.  \frac{1}{2}

E.  \frac{47}{80}

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harshittahuja · Answer

Three friends play marbles each week. When they combine their marbles, they have 100 in total. 45 of the marbles are new and the rest are old. 30 are red, 20 are green, 25 are yellow, and the rest are white. What is the probability that a randomly chosen marble is new OR yellow?
Solution:-
Total=100
new marbles are =45
Old ones are =55
out of total 
red=30
green = 20
Yellow=25 
White= 25 
prob that new is =0.45
prob that yellow=0.25
case when a marble is yellow and New=0.25*0.45
So
0.45+0.25-0.1125
=0.70-0.1125
=0.5875
so answer can be =E 47/80

Regor60 · Answer

Probability of new or yellow is:

1-probability of neither

Probability of not yellow:

75/100

Probability of not new (old):

55/100

Probability of neither:

(55*75)/(100)(100) =

11/20 * 3/4 = 33/80

1- 33/80 = 47/80

Posted from my mobile device

jaegarbong · Answer

Correct me if I am wrong here.

marble is new  OR   yellow?

If the wording is OR - P(New) + P(Yellow)

If the wording is AND -> P(new) * P(Yellow)

However, I recollect a formula - P(A or B) = P(A) + P(B) - P(A & B) 
We dont have P(A&B) here for correct calculation.

Regor60 · Answer

You're correct and the only way to arrive at an answer is to assume independence, which is unsatisfying and a slippery slope in fostering unwarranted assumptions as to independence generally

haianh · Answer

I arrived at a different answer for this problem and would really appreciate if someone could confirm whether my reasoning is incorrect. In cases like this, can we generally assume that P(A and B) = P(A) x P(B) if no additional information is provided?

We are given:  
 
  Total marbles: 100.  
  Red = 30, Green = 20, Yellow = 25, White = 25  
  New = 45, Old = 55.   
 Task:  find the probability a randomly chosen marble is New or Yellow = P(n or y).
 
 P(n or y) = P(n) + P(y) - P(n&y) = 1/4 + 9/20 - P(n&y) = 7/10 - P(n&y) 
 P(n&y) min = 0 
 P(n&y) max = 25/100 = 1/4  
 So:   \frac{9}{20} \leq  P(n or y)  \leq \frac{7}{10}

We also know that marble number is an integer, so the no. of new yellow marbles must also be an integer   	o  P(n&y)*100 must be an integer.
Now we check the answer choices:
A. 4/53 (too small)  	o  Out
B: 3/13 (too small)  	o  Out
C: 7/10 - 7/15 = 7/30. 7/30*100 is not an integer  	o  Out
D: 7/10 - 1/2 = 1/5. 1/5*100 is an integer  	o  Keep
E: 7/10 - 47/80 = 9/80. 9/80*100 is not an integer  	o  Out

Most probable answer: D

DonBosco7 · Answer

Three friends play marbles each week. When they combine their marbles, they have 100 in total. 45 of the marbles are new and the rest are old. 30 are red, 20 are green, 25 are yellow, and the rest are white. What is the probability that a randomly chosen marble is new OR yellow?

solution:-

P (New or Yellow) = P(New) + P(Yellow) - P(New and Yellow)
= 45/100 + 25/100 - (45/100 * 25/100)
= 47/80

option E

solution:-

P (New or Yellow) = P(New) + P(Yellow) - P(New and Yellow)
 = 45/100 + 25/100 - (45/100 * 25/100)
 = 47/80

option E

Three friends play marbles each week. When they combine their marbles,

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Three friends play marbles each week. When they combine their marbles,

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