Wiki User
∙ 13y agoat least 39 blue marbles
Wiki User
∙ 13y ago48 marbles
Set up the problem as: y=number of yellow marbles & 4y=number of blue marbles. 40=y+4y or 40=5y or y=8. Therefore number of blue marbles is 4*8 = 32.
12 blue marbles
If there are 20 red marbles and 40 blue marbles, the ratio of red to blue is 1 over 2. The number of white marbles does not matter.
It is 1248/57120 = 13/595 = 0.0218 approx.
10 Green marbles, 13 Blue marbles.
There are at least 11 green marbles in the bag.
He will have 13 blue marbles and 10 green marbles.
He has 10 green marbles.
Depends on the color of the marbles, or more specifically, it depends upon how many of the marbles are blue. If 0 marbles are blue, then the chance is 0. If 1 marble is blue, then the chance is 1/4. If 2 marbles are blue, then the chance is 1/2. If 3 marbles are blue, then the chance is 3/4. If 4 marbles are blue, then the chance is 1. These probabilities assume that you draw only once. The other possibilities are that you draw twice, or three times, or four times. If drawing multiple times, it can be done by putting back each marble after it was drawn, or by keeping it out after it has been drawn.
6
48 marbles
Set up the problem as: y=number of yellow marbles & 4y=number of blue marbles. 40=y+4y or 40=5y or y=8. Therefore number of blue marbles is 4*8 = 32.
1 in 5.45 of picking at least 2 blue marbles any 2 from 3 is 3 correct picks =3X7(red)=21 of 2 blue only 1 pick of 3 blue only any 3 from 10=120 120/22=5.45
a+b=23 b=a+3 a+b a+(a+3)=23 a+a=23-3 2a=20 a=10 green marbles=10 blue marbles =23-10 =13
Let X = the number of green marbles. X+3 = the number of blue marbles. X + (X+3) = 23 2X + 3 = 23 2X = 20 X = 10 or the number of green marbles.
12 blue marbles