there are six red marbles 4 green and 2 white marbles in a box what is the probability of drawing a non white marble ? Koi h tya ? About the author Eva
[tex]\large\green{\bold{Given:}}[/tex] There are six red marbles 4 green and 2 white marbles in a box. [tex]\large\green{\bold{To \: find}}[/tex] What is the probability of the drawing a non white marble? [tex]\large\green{\bold{Solution}}[/tex] Total numbers of marbles = Red marbles + Green marbles + White marbles = 6 + 4 + 2 = 12 [tex]\large\green{\bold{As \: we \: know \: that, }}[/tex] ★[tex]\large \: probabiliy = \frac{favourable \: outcome }{total \: number \: of \: outcome} [/tex] Consider probability of getting non white be P(E) Non white marble = Green marble + red marble = 6 + 4 = 0 [tex] \implies[/tex]P(E) = [tex]\large \frac{10}{12} [/tex] = [tex]\large \frac{5}{6} [/tex] .·. The probability of getting non white marble is [tex]\large \frac{5}{6} [/tex] Reply
Step-by-step explanation: P(Both Red) + P(Second is Red): P(Both Red) =(22)(92) P(Second is Red) =(71)(21)(92) This comes out to be 512. Reply
[tex]\large\green{\bold{Given:}}[/tex]
There are six red marbles 4 green and 2 white marbles in a box.
[tex]\large\green{\bold{To \: find}}[/tex]
What is the probability of the drawing a non white marble?
[tex]\large\green{\bold{Solution}}[/tex]
Total numbers of marbles = Red marbles + Green marbles + White marbles = 6 + 4 + 2 = 12
★[tex]\large \: probabiliy = \frac{favourable \: outcome }{total \: number \: of \: outcome} [/tex]
Consider probability of getting non white be P(E)
[tex] \implies[/tex]P(E) = [tex]\large \frac{10}{12} [/tex] = [tex]\large \frac{5}{6} [/tex]
.·. The probability of getting non white marble is [tex]\large \frac{5}{6} [/tex]
Step-by-step explanation:
P(Both Red) + P(Second is Red):
P(Both Red) =(22)(92)
P(Second is Red) =(71)(21)(92)
This comes out to be 512.