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anonymous
 4 years ago
Please see the attachment and answer with explanation.
anonymous
 4 years ago
Please see the attachment and answer with explanation.

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Please see these and answer.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0haha thats like a whole assignment on probability please post one question at a time

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0lol..yeahh..but they r tiny ones...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0At least solve questions 1, 2 , 3.

vishal_kothari
 4 years ago
Best ResponseYou've already chosen the best response.4the link is not opening in my computer..so draw the questions...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Question 1 ) From a group of 3 boys and 2 girls, 2 children are selected at random. Find the Sample Space.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Options nahin hain. Model paper for SA2 question.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Question 2) numbers are thrown getting 2 numbers whose sum is divisible by 4 or 5, is considered as a success. Find the probability og getting success.

vishal_kothari
 4 years ago
Best ResponseYou've already chosen the best response.4what is tough in that...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Question1) answer is 20. I guess.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Question 3) A coin is tossed 3 times. Find the chance that head and tail show alternately. (I think answer is 4/8 = 1/2) Plz verify and say.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ok, I'll take a crack at this for you. So, for #1, the Sample Space is defined as the list of all possible outcomes. The question is, is 1 boy and 1 girl the option? Or are they unique, like Boy 1 and Girl 1 vs. Boy 1 and Girl 2  does that make sense? So, I'm not 100% on how the question is defined to be honest... #2: You need to determine all the possible combinations of the sum of two die (2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12), see the ones divisible by 4 or 5, then divide that number by they number of possibilities (in this example there are 11). #3: A coin flip is a 50/50 chance. Alternatively, this written as 1/2 (1 possible successful outcome out of two possible outcomes). So, when flipping a coin three times, you need to multiply the respective probabilities to find your answer. Hope this helps

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Can u please explain questions 1 and 3 in detail? The detailed solution.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0For #1 I really can't  the problem doesn't give enough parameters for me to be sure. The issue is how they define the Sample Space. If one boy and one girl is a possible outcome, is it just ONE possible outcome? Does it matter WHICH boy or girl are chosen? Also, does the order matter? Is one boy and one girl the same or different from one girl and one boy? #3 What exactly are you looking for here? Which part of the explanation do you need help with?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0In question 1, waht matters is the grouping I think. Means Boyboy, boy0girl, and so on. In question 3, The sample space is: (HHH), (HHT), (HTH), (HTT), (TTT), (TTH), (THT) and (THH). So I think probability of getting head and tail alternately is 4/8 = 1/2.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0For #1, ok, so in that case it would be 20 then? For #3, aren't there only two correct outcomes? THT & HTH?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0THen ok. For no.3, I am doubt. I think u r correct.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0And for #1  my concern is this, here are the two sample spaces which I can't decide between: BG, BB, GG, GB OR B1G1, B1G2, B2G1, B2G2, B3G1, B3G2, and so on...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I tried with first assumption though I am not sure.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ok sweet, FFM is here  I really want to see his opinion on this

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0>> From a group of 3 boys and 2 girls, 2 children are selected at random. Find the Sample Space. I will just answer the number in the sample space: \( \binom 32 +\binom 22+ \binom 31\times \binom21 =\cdots\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So the final answer? I don't know Permutation and combination those things.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0How could you possible understand probability without understanding elementary combinatorics?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Those things are not there in our syllabus. We just find the sample space by counting. That's the scope of syllabus till now.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Hm fair enough, Good luck :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dat was answer for 1st ques..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0But B1G1, B2G2, that can also come na? I did that way. Salini didi said that.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yeah ...u r right....hwmany marks ques r r these??

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0itz....bttr if u write ...as u have written

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it doesn't indicate that it matters which boy or which girl is selected 4 possibilities GG GB BG BB

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yeah dumbow is also right...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and if it did then it would have 20 possibilities 5P2 = 20

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I also got 20 but............

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0c the ques . is 2 children are selected ....and children is classified as male female ...so ans is BB GG BG gb

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0children are generally clasified as boy girl onle..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0for #2 does it mean 2 dice are thrown ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oops srry mate ...i ead it as dice....

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0did u take 0's into account?/

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0only numbers divisible be 4or5> 4,5,8,10,12 3 ways of rolling a 4 4 ways of rolling a 5 5 ways of rolling a 8 3 ways of rolling a 10 1 way of rolling a 12 > 16/36 = 4/9

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i said na ..its not a dice,.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0nope...and second ans is 16/36=4/9
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