3x + (x+3) = 19
4x + 3 = 19
4x = 16
x = 4
y= x+3
y= 4+3
y= 7
Complete question:
Consider 3 boxes, each of which contains 4 balls in particular, box 1 contains 4 white balls, box 2 contains 3 white balls and 1 red ball, box 3 contains 2 white balls and 2 red balls. Frank chooses a box at random. If boxes 1 and 3 are chosen, Frank extracts 2 balls without replacement. If box 2 is chosen Frank extracts 2 balls with replacement. USE ONLY THE CONDITIONAL PROBABILITY FORMALISM and derive an expression for the probability that the 2 extracted balls are red.
Answer:
11/288
Step-by-step explanation:
We are given:
Box 1: ( 4White, ORed)
Box 2: (3White, 1Red)
Box 3: (2White, 2 Red)
We are told that if Frank choose from Box 1 and Box 3, 2 balls are extracted without replacement.
Since there is no red ball in Box 1, there is no way 2 red balls will come out from Box1.
Our Event, E = getting 2 red balls.
Now Box 1 is ruled out, we have:
P[E(B1)]= 0
P[E/B3)] = (2 2) / (4 2)
= 1/6
If box 2 is chosen, 2 balls are extracted with replacement. Therefore for Box 2:
P(E/B2) = (1/4) *(1/4)
= 1/16
Therefore probability that 2 balls extracted are red, we have:
P(E)=P(E/B1) P(B1) + P(E/B2) P(B2)+P(E/B3) P(B3)

= 11/288
The representative sample would be expected to have 6 blue marbles 8/24:2/x
Answer:
y^ -2
Step-by-step explanation:
We know that a^b^c = a^(b*c)
(y^-1/2)^4
y^(-1/2*4)
y^ -2