1. Box A has three cells labeled 1 to 3. Box B has four cells labeled 1 to 4. Th
ID: 3069712 • Letter: 1
Question
1. Box A has three cells labeled 1 to 3. Box B has four cells labeled 1 to 4. There are three yellow balls numbered 1 to 3, and four green balls numbered 1 to 4. The yellow balls are placed at random in Box A, and the green balls are placed at random in box B, with no cell receiving more than one ball. Find the probability that only one of the boxes will show no matches. (A match occurs when the cell's number matches the number on the ball in it.) 2. Let {Snin 1} be a sequence of-algebras. Show that nn. 9, is a-algebra.Explanation / Answer
1.
Box A has 3 balls
Now the ways of placing 1,2,3 numbered box in box A are 3!
while the no match will occur for 2 times (231,321 for a seq of 123)
So the prob of no match for A =2/3! = 2/6=1/3
Same way for Box B
the ways of placing 1,2,3,4 numbered box in box B are 4!
while the no match will occur for 9 times (i.e. 2341,2413,2143,3142,3421,3412,4123,4312,4321 for a seq of 1234)
So the prob of no match for B =9/4! =9/24= 3/8
Now we need to find the prob of only one of the boxes shows no matches
Which is the.,
P(Box A show no match)*P(Box B Doesn't Show no match) + P(Box A doesn't show no match)*P(Box B Show no match)
=1/3*(1-3/8) + (1-1/3) * 3/8 = 0.4583333
So., 0.4583333 is the prob of only one of the boxes shows no matches
Hope the above answer has helped you in understanding the proble. Please upvote the ans if it has really helped you. Good Luck!!
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