Can anyone please help me
2 answers:
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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
Answer:
Choice A:
Step-by-step explanation:
means we looking for first term 5 and the sequence is going up by 2.
In general,
means you have first term and the sequence has a common difference of d.
So it is between the first two choices.
The explicit form of an arithmetic sequence is:
An equivalent recursive form is
So d again here is 2.
So choice a is correct.