SOLUTION: 1. From a box containing 6 red bulbs, 8 yellow bulbs and blue bulbs, two bulbs is drawn. Dteremine the probability that the first is a red and the second is yellow.
2. Five gir
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Question 47810: 1. From a box containing 6 red bulbs, 8 yellow bulbs and blue bulbs, two bulbs is drawn. Dteremine the probability that the first is a red and the second is yellow.
2. Five girls and 3 boys are to be seated in a row. If seats are taken randomly, waht is the probability that the girls will be seated together?
3. A box of 15 batteries contains defectives, what is the probability that out of the 3 baterries selected without replacement, two is not defective?
4. A box contains 5 white and 3 green marbles. A second box contains 4 white and 6 green marbles. Suppose that a marble is drawn from the first box and placed unseen in the second box. What is the probability that a marble now drawn from the second box is green?
Found 2 solutions by consc198, venugopalramana:
Answer by consc198(59) (Show Source): You can put this solution on YOUR website!
1. 1/9
2. 1/16
3. 1/15
4. 9/18 = 1/2
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
1. From a box containing 6 red bulbs, 8 yellow bulbs and (??????HOW MANY )blue bulbs, two bulbs is drawn. Dteremine the probability that the first is a red and the second is yellow.
2. Five girls and 3 boys are to be seated in a row. If seats are taken randomly, waht is the probability that the girls will be seated together?
TOTAL NUMBER OF POSSIBILITIES TO SEAT 8 PERSONS=8!
FOR THE 3 GIRLS TO BE SEATED TOGETHER,CONSIDER 3 GIRLS TOGETHER AS ONE UNIT..THEN TOTAL =5+1=6
THEY CAN BE SEATED IN 6! WAYS
HENCE P=6!/8!=1/(8*7)=1/56
3. A box of 15 batteries contains( ?????HOW MANY ??)defectives, what is the probability that out of the 3 baterries selected without replacement, two is not defective?
4. A box contains 5 white and 3 green marbles. A second box contains 4 white and 6 green marbles. Suppose that a marble is drawn from the first box and placed unseen in the second box. What is the probability that a marble now drawn from the second box is green?
CASE 1....LET A GREEN MARBLE BE DRAWN FROM BOX 1
ITS PROBABILITY = 3/8
NOW WHEN IT IS PUT IN THE SECOND BOX THERE ARE 4 WHITE AND 7 GREEN MARBLES
P OF DRAWING GREEN MARBLE = 7/11
SO COMBINED PROBABILITY = (3/8)(7/11)=21/88
CASE 2......LET A WHITE MARBLE BE PICKED UP FROM BOX 1
ITS PROBABILITY = 5/8
NOW WHEN IT IS PUT IN THE SECOND BOX THERE ARE 5 WHITE AND 6 GREEN MARBLES
P OF DRAWING GREEN MARBLE = 6/11
SO COMBINED PROBABILITY = (5/8)(6/11)=30/88
HENCE P OF EITHER HAPPENING = 21/88 + 30/88 = 51/88
You may edit the question. Maybe convert formulae
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