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grade 8 probability questions and answers
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Question:please solve and show set up...best answer to whoever does BOTH
the weather forecaster has predicted there is a 25% probability of rain on any of the next 4 days. what is the probability that it will rain at least twice?
a) 66/256
b) 67/256
c) 68/256
d) 175/256
Answers:P(rain on any ONE given day) = 25% = P(rain ONLY once over a 4day period) = P(rain on 1st day OR rain on 2nd day OR rain on 3rd day OR rain on last day) = ( x x x ) + ( x x x ) + ( x x x ) + ( x x x ) = ( x x x ) X 4 = 108/256 = 0.421875 = 42.2% "P(rain at least twice)"? = P(rain more than once) = 1 P(rain ONLY once over a 4day period) = 1 108/256 = 148/256 = 0.578125 = 57.8% I do not see this answer in the choices you offered.
Answers:P(rain on any ONE given day) = 25% = P(rain ONLY once over a 4day period) = P(rain on 1st day OR rain on 2nd day OR rain on 3rd day OR rain on last day) = ( x x x ) + ( x x x ) + ( x x x ) + ( x x x ) = ( x x x ) X 4 = 108/256 = 0.421875 = 42.2% "P(rain at least twice)"? = P(rain more than once) = 1 P(rain ONLY once over a 4day period) = 1 108/256 = 148/256 = 0.578125 = 57.8% I do not see this answer in the choices you offered.
Question:I need to know how to get an answer to this question.Okay, suppose you roll a six faced die 600 times. About how many times would you expect to get a 4? An odd number?
Answers:100 times 300 times.
Answers:100 times 300 times.
Question:we did a math test today.. and the teacher said that most of us got one page alll wrong.
i dont think i did though? idk
can you help me with these questions??
1) a family has 4 pets. some are dogs. some are cats.
what is the probibility of:
A) 2 dogs and 2 cats
B) more dogs than cats
C) more than 2 cats
2) there are 4 people in a race. Matt, Bill, Tina, Jen
what are the possible combinations of the 1st two people that win the race?
what is the probabilty of:
A) both girls beat the boys.
B) tina wins the race these were my answers....
for the first question... i got....
since it said that SOME are dogs and SOME are cats.. it means theres a mix of both, and you can have all dogs or all cats..
so the only combinations were..
1 dog, 3 cats
2 dogs, 2 cats
3 dogs, 1 cat.
A) 1/3 B) 1/3 C) 1/3
for the second question.. i got....
since it only said the first 2.. you only need to get combos for the 1st 2 ppl.. right?
MB BM TM JM
MT BT TB JB
MJ BJ TJ JT
12 difference outcomes.
A) 2/12  1/6 B) 3/12  1/4
Answers:1) A family has 4 pets, some are dogs, some are cats. What is the probability of: A) 2 dogs and 2 cats B) more dogs than cats C) more than 2 cats Interesting question.  It depends on *exactly* how the question was worded. If the question reads, "some are dogs, some are cats", then the ONLY probability is: A) Two dogs and two cats = 100% (Dogs and Cats are plural). However, I believe the question most likely read like this: 1) A family has four pets. The animals can be dogs or cats. What is the probability of: A) 2 dogs and 2 cats B) more dogs than cats C) more than 2 cats A) 2 Dogs and 2 Cats P(2 Dogs, 2 Cats) = 4! / (2! * 2!) * (1/2)^2 * (1/2)^2 = 3/8 B) More Dogs than Cats P(Dogs > Cats) = P(3 Dogs, 1 Cat) + P(4 Dogs) P(Dogs > Cats) = [ 4! / (3! * 1!) * (1/2)^3 * (1/2)^1 ] + (1/2)^4 P(Dogs > Cats) = [ 1/4 ] + (1/16) = 5/16 C) More than 2 Cats P(Cats > 2) = P(3 Cats, 1 Dogs) + P(4 Cats) P(Cats > 2) = [ 4! / (3! * 1!) * (1/2)^3 * (1/2)^1 ] + (1/2)^4 P(Cats > 2) = [ 1/4 ] + (1/16) = 5/16 The above computations are using the BINOMIAL FORMULA. P(k) = N! / [k! * (N  k)!] * p^k * q^(N  k) where: N = Number of opportunities for event x to occur (4) k = Number of times that event x should happen (3) p = Probability of a success (1/2) q = Probability of failure (1/2) ! = Factorial  2) There are 4 people in a race. Matt, Bill, Tina and Jen. What are the possible combinations of the 1st two people that win the race? What is the probability of: A) Both girls beat the boys. B) Tina wins the race. There are 24 possible combinations, (4! = 24) Only four combinations where both girls beat the boys: T J M B T J B M J T M B J T B M Therefore, the probability that both girls beat both boys equals 4/24 or 1/6. B) Probability that Tina wins the race: Short answer = one chance out of four or 1/4. Long answer: There are only six combinations for Tina to win: T B J M T B M J T M J B T M B J T J M B T J B M 6 / 4! = 6/24 = 1/4 Good luck in your studies, ~ Mitch ~
Answers:1) A family has 4 pets, some are dogs, some are cats. What is the probability of: A) 2 dogs and 2 cats B) more dogs than cats C) more than 2 cats Interesting question.  It depends on *exactly* how the question was worded. If the question reads, "some are dogs, some are cats", then the ONLY probability is: A) Two dogs and two cats = 100% (Dogs and Cats are plural). However, I believe the question most likely read like this: 1) A family has four pets. The animals can be dogs or cats. What is the probability of: A) 2 dogs and 2 cats B) more dogs than cats C) more than 2 cats A) 2 Dogs and 2 Cats P(2 Dogs, 2 Cats) = 4! / (2! * 2!) * (1/2)^2 * (1/2)^2 = 3/8 B) More Dogs than Cats P(Dogs > Cats) = P(3 Dogs, 1 Cat) + P(4 Dogs) P(Dogs > Cats) = [ 4! / (3! * 1!) * (1/2)^3 * (1/2)^1 ] + (1/2)^4 P(Dogs > Cats) = [ 1/4 ] + (1/16) = 5/16 C) More than 2 Cats P(Cats > 2) = P(3 Cats, 1 Dogs) + P(4 Cats) P(Cats > 2) = [ 4! / (3! * 1!) * (1/2)^3 * (1/2)^1 ] + (1/2)^4 P(Cats > 2) = [ 1/4 ] + (1/16) = 5/16 The above computations are using the BINOMIAL FORMULA. P(k) = N! / [k! * (N  k)!] * p^k * q^(N  k) where: N = Number of opportunities for event x to occur (4) k = Number of times that event x should happen (3) p = Probability of a success (1/2) q = Probability of failure (1/2) ! = Factorial  2) There are 4 people in a race. Matt, Bill, Tina and Jen. What are the possible combinations of the 1st two people that win the race? What is the probability of: A) Both girls beat the boys. B) Tina wins the race. There are 24 possible combinations, (4! = 24) Only four combinations where both girls beat the boys: T J M B T J B M J T M B J T B M Therefore, the probability that both girls beat both boys equals 4/24 or 1/6. B) Probability that Tina wins the race: Short answer = one chance out of four or 1/4. Long answer: There are only six combinations for Tina to win: T B J M T B M J T M J B T M B J T J M B T J B M 6 / 4! = 6/24 = 1/4 Good luck in your studies, ~ Mitch ~
Question:Attending a lesson = 0.7  Improving = 0.9  Not improving = o.1
Not attending a lesson = 0.3  Improving = 0.4  not improving = 0.6
A person improves
Whats the probability he attends? The answer in 0.84 but how do you do it! Please help!
Thanks so much in advance :)
Answers:tree diagram ==========. ................................ improves 0.9 > 0.7*0.9 = 0.63 = A ............ attends 0.7 ................................. doesn't 0.1 XX student ................................. improves 0.4 > 0.3*0.4 = 0.12 = B ............. doesn't 0.3 .................................. doesn't 0.6 XX P[attends & improves] = 0.7*0.9 = 0.63 = A P[doesn't attend & improves] = 0.3*0.4 = 0.12 = B P[improves] = A+B = 0.75 P[attends  improves] = A/(A+B) = 0.84 <
Answers:tree diagram ==========. ................................ improves 0.9 > 0.7*0.9 = 0.63 = A ............ attends 0.7 ................................. doesn't 0.1 XX student ................................. improves 0.4 > 0.3*0.4 = 0.12 = B ............. doesn't 0.3 .................................. doesn't 0.6 XX P[attends & improves] = 0.7*0.9 = 0.63 = A P[doesn't attend & improves] = 0.3*0.4 = 0.12 = B P[improves] = A+B = 0.75 P[attends  improves] = A/(A+B) = 0.84 <