SOLUTION: in a group of 25 people, 19 are right handed 4 are left- handed and 2 is ambidextrous if 2 people are selected at random without replacement use a tree to find the following prob

Question 1126605: in a group of 25 people, 19 are right handed 4 are left- handed and 2 is ambidextrous if 2
people are selected at random without replacement use a tree to find the following probabilities

a) both are left handed
b) one is right-handed,and one is left handed
c) two are right-handed
d) two are ambidextrous
Found 2 solutions by greenestamps, Alan3354:
Answer by greenestamps(13198) (Show Source): You can put this solution on YOUR website!

Trees don't work very well in typed text; I'll let you draw the tree.

a) P(both left) = P(first left AND second left) = (4/25)(3/24) = 1/50

b) P(one each) = P(first left AND second right OR first right AND second left) = P(first left AND second right) + P(first right AND second left) = (4/25)(19/24)+(19/25)(4/24) = 19/150+19/150 = 19/75

c) P(both right) = P(first right AND second right) = (19/25)(18/24) = 57/100

d) P(both ambidextrous) = P(first ambidextrous AND second ambidextrous) = (2/25)(1/24) = 1/300
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
I'd give my right arm to be ambidextrous.

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