a set contains the numbers 0,6,12, and 15 Two different numbers are selected randomly from this set. What is the probability tha
1 answer:
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Solution:
Data represented as number of hours spent in studying by Group A Students:
1,2,1,1,3,3,2,2,3
Arranging it in ascending order: 1,1,1 ,2, 2, 2, 3, 3, 3,
As number of terms is odd, The median will be middle value of observation.Which is 2.
The Data arranged in ascending order are , (1,1,1,2),2(2,3,3,3).
Median of (1,1,1,2) = =1
Median of (2,3,3,3)==3
=Interquartile Range =
=3-1=2
For Data Set 2,
The Data for group B students are: 3 2 3 2 2 2 1 1 2
Arranging in ascending order: 1,1,2,2,2,2,2,3,3
total number of observation = 9
Median = 2
Arranging the data as : (1,1,2,2) 2,(2,2,3,3)
Median of (1,1,2,2)= Number of observation is 4 which is even , so Median= =
Median of (2,2,3,3)==
S=Interquartile Range = =
Interquartile range for Group A Students =Interquartile range for Group B students + 1
Option (D) The interquartile range for Group A employees is 1 more than the interquartile range for Group B students is true.
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