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 3 solutions by consc198, venugopalramana, ikleyn:
Answer by consc198(59)   (Show Source):
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
Answer by ikleyn(53937)  (Show Source):
You can put this solution on YOUR website! .
It is absolutely inappropriate way to communicate with the tutors by posting
4 (four) different Math problems in one post.
The rules of this forum (and the rules of any other similar forum) require
ONE and ONLY ONE problem per post.
Why such rules are established ? - - - Due to two reasons.
First reason is the requirement of polite communication.
When you place more than one problem per post, you demonstrate disrespect to tutors.
Second reason is that when you place many assignments in one post, the result is a mess,
which is difficult to write and difficult to read.
Simply separate your post in 4 (four) different posts, and you will get
perfect solutions with detailed and complete explanations !
Always be polite with the tutors - and happiness will be yours !
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