roblein A family consisting of 2 parents, 4 children, and 1 dog is taking a pict
ID: 3041025 • Letter: R
Question
roblein A family consisting of 2 parents, 4 children, and 1 dog is taking a picture with all 7 family members standing in a row (a) Find the probability that the two parents are not next to each other. (b) Find the probability that the dog is standing somewhere to the right of all 4 childrern. Problem 2 In a group of 200 students, 73 speak Spanish, 63 speak Italian, and 56 speak French. Additionally, 22 speak Spanish and Italian, 18 speak Spanish and French, 15 speak Italian and French, and 5 speak all three Find the probability that a randomly chosen student speaks exactly one of the three languages. Problem 3 A standard 52-card deck contains 13 ranks (A. 2. 3. 4. 5, 6, 7, 8, 9, 10. J. Q. K) and 4 suits (.: ): (a) Find the probability that a 5 card hand is missing at least one suit example: 29, 4 example: 29, 4 . J 10 . K (missing 4) , J 10 . K (missing ..) (b) Find the probability that a 10-card hand contains one 3-of-a-kind, three pairs, and one single (for a total of five different ranks)Explanation / Answer
Solution:-
1)
a) The probability that two parents are not next to each other is 0.8571.
Total number of ways of sitting = 7! = 5040
Number of ways two parents sits together = 6! = 720
Number of ways two parents does not sits together = 5040 - 720 = 4320
The probability that two parents are not next to each other = 4320/5040 = 0.8571
b) The probability that dog is standing somewhere to the right of all 4 childrens is 0.20.
Total number of ways of sitting = 7! = 5040
Number of ways dog at right corner = 6! = 720
Number of ways dog at one seat before right corner = 2 × 5! = 240
Number of ways dog at two seat before right corner = 4! × 2! = 48
Number of ways dog is standing somewhere to the right of all 4 childrens = 720 + 240 + 48 = 1008
The probability that dog is standing somewhere to the right of all 4 childrens = 1008/5040 = 0.20
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