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anonymous
 5 years ago
A fair die is tossed 4 times. Find the probability of a. at least 2 5s b. at most 3 even numbers c. no more than 1 odd number d. at least 1 prime number e. at most 1 number greater than 4
anonymous
 5 years ago
A fair die is tossed 4 times. Find the probability of a. at least 2 5s b. at most 3 even numbers c. no more than 1 odd number d. at least 1 prime number e. at most 1 number greater than 4

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you use the fact that P(A) = 1 P(not A)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0can u show me how to do each part?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0P(at least two 5s) = 1 P(not at least two 5s)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0= 1 P(at most one 5)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so let's find P(at most one 5)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I am not finding the answer, just trying to help

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so P(at most one 5) = P(no 5s) + P(exactly one 5)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0P(no 5) = \[(5/6)^4\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0P(exactly one 5) follows the binomial distribution, so \[4C1*(1/5)^1*(5/6)^4\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0when you subtract the sum of those from 1, you get the answer for a),

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oops, the 1/5 should be 1/6

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0don't u need 0 5s as well?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I already did that, it is (5/6)^4

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wait question is at least 25s

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0let me explain you that par again

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0P( at least two 5s) = P(exactly two) + P(exactly three)+P(exactly four)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it is a pain to calculate all those, so we use the rule P(event) = 1P(complement of that event)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.04C2(1/6)^2(5/6)^3 + 4C3(1/6)^3(5/6)^1 + 4C4(1/6)^4(5/6)^0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what I did was 1 (5/6)^6 4C1(1/6)(5/6)^3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0for B, the complement is "no evens at all"

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so 1P(no evens) will be the answer

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0isn't it \[4C3(1/2)^3(1/2)^1 + 4C2(1/2)^2(1/2)^2 + 4C1 (1/2)^1(1/2)^3\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you need to include 4C0(1/2)^4

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0everything else is correct?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0az, just to let you know you probably know how to get the answers for all of this. But you also need to know that you are missing the point of what these problems are asking.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so it can be 0 or 1 odd number?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.04C1(1/2)^1 (1/2)^3 + 4C0(1/2)^0 (1/2)^4 ??

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0az, I need to go soon so let me leave you an advice

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0your method is logically straight forward, and as long as you don't mind calculating a lot of numbers you will get the answer so I have no doubt that you will get any of these wrong from what I have seen so far

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0can we go over the last two really quickly

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but all of the problems here can be found a lot more easily using the fact P(A) = 1P(notA)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.04C1 (1/2)^1(1/2)^3 + 4C0 (1/2)^0 (1/2)^4 ????

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so d is 1  P (all of them being prime) and e is P(0) + P(1)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is that right? what I just put

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry, ask someone else since I need to go. :)
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