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haha thats like a whole assignment on probability please post one question at a time
lol..yeahh..but they r tiny ones...
At least solve questions 1, 2 , 3.
the link is not opening in my computer..so draw the questions...
Question 1 ) From a group of 3 boys and 2 girls, 2 children are selected at random. Find the Sample Space.
Options nahin hain. Model paper for SA-2 question.
Question 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.
what is tough in that...
Question1) answer is 20. I guess.
Question 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.
Ok, 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
Can u please explain questions 1 and 3 in detail? The detailed solution.
For #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?
In question 1, waht matters is the grouping I think. Means Boy-boy, 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.
For #1, ok, so in that case it would be 20 then? For #3, aren't there only two correct outcomes? THT & HTH?
THen ok. For no.3, I am doubt. I think u r correct.
And 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...
I tried with first assumption though I am not sure.
Ok sweet, FFM is here - I really want to see his opinion on this
>> 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: \( \binom 32 +\binom 22+ \binom 31\times \binom21 =\cdots\)
So the final answer? I don't know Permutation and combination those things.
How could you possible understand probability without understanding elementary combinatorics?
Those things are not there in our syllabus. We just find the sample space by counting. That's the scope of syllabus till now.
Hm fair enough, Good luck :)
dat was answer for 1st ques..
But B1G1, B2G2, that can also come na? I did that way. Salini didi said that.
yeah ...u r right....hwmany marks ques r r these??
itz....bttr if u write ...as u have written
it doesn't indicate that it matters which boy or which girl is selected 4 possibilities GG GB BG BB
yeah dumbow is also right...
and if it did then it would have 20 possibilities 5P2 = 20
I also got 20 but............
c the ques . is 2 children are selected ....and children is classified as male female ...so ans is BB GG BG gb
children are generally clasified as boy girl onle..
3 ans ..... 2/8 =1/4
for #2 does it mean 2 dice are thrown ?
oops srry mate ...i ead it as dice....
i get 4/9 for #2
did u take 0's into account?/
only 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
i said na ..its not a dice,.
nope...and second ans is 16/36=4/9
Thanks a lot