If 'a' is an odd natural number and 'b' is an even natural number, what is the total number of solutions of theequation ab = 2a + b + 598?A. 16B. 12C. 5D. 6E. 8
A coin is biased in such a manner that the chances of getting head is twice the chances of getting tail.Find the probability that the heads will appear an odd number of times in n tosses of the coin.
A coin is biased in such a manner that the chances of getting head is twice the chances of getting tail.Find the probability that the heads will appear an odd number of times in n tosses of the coin.
bhai dont hav d options and OA .......just seen dis question somewhere......thought it is a worth sharing question........with the hope of gaining some knowledge.....plz at least share ur thoughts on dis....
A coin is biased in such a manner that the chances of getting head is twice the chances of getting tail.Find the probability that the heads will appear an odd number of times in n tosses of the coin.
nC1*p^1*q^(n-1) + nC3*p^3*q^(n-3) + ....nCn*p^n*q^0 ---If n is odd
A coin is biased in such a manner that the chances of getting head is twice the chances of getting tail.Find the probability that the heads will appear an odd number of times in n tosses of the coin.
If 'a' is an odd natural number and 'b' is an even natural number, what is the total number of solutions of theequation ab = 2a + b + 598?A. 16B. 12C. 5D. 6E. 8 (2
ab = 2a + b + 598
=> (a-1)*(b-2) = 600 = 2^3*3*5^2
Here, a = odd and b = even
So, a-1 = even and b-2 = even
Hence, Both will have a factor of 2
Therefore 2^2 is common on both sides and can be striked out.
nC1*p^1*q^(n-1) + nC3*p^3*q^(n-3) + ....nCn*p^n*q^0 ---If n is oddp= 1/3 , q =2/3 ??
My interpretation of your solution here you have taken n as odd also q is prob of head and p is prob of tail Now if n is odd and every time in q's power (odd- odd)=even.......is reflecting the probability of head appearing even times(in question it has been asked for odd times) One more thing ,don't you think one more case, when n is even is also required(as n is number of tosses,it can be either even or odd) PS :dont hav d OA ....so not sure about any of my claims(just interested in some hard core discussion)
My interpretation of your solutionhere you have taken n as oddalso q is prob of head and p is prob of tailNow if n is odd and every time in q's power (odd- odd)=even.......is reflecting the probability of head appearing even times(in question it has been asked for odd times)One more thing ,don't you think one more case, when n is even is also required(as n is number of tosses,it can be either even or odd)PS ont hav d OA ....so not sure about any of my claims(just interested in some hard core discussion)
Firstly, the question is the probability of finding odd no of heads in "n" tosses.
So, Suppose n=odd (I know it can be even, the equation won't be much different)
Then, no of heads can be = 0, 1, 2, 3, 4,. ....n..
No of heads odd = 1, 3, 5, .....n (If n=odd)
= 1, 3, 5, ...(n-1) --(If n= even)
So, No of ways of 1 head is = nC1*p^1*q^(n-1)
Similarly for others
Final equation would come out to be the one posted earlier.. :)
Firstly, the question is the probability of finding odd no of heads in "n" tosses.So, Suppose n=odd (I know it can be even, the equation won't be much different) Then, no of heads can be = 0, 1, 2, 3, 4,. ....n..No of heads odd = 1, 3, 5, .....n (If n=odd)= 1, 3, 5, ...(n-1) --(If n= even)So, No of ways of 1 head is = nC1*p^1*q^(n-1) Similarly for othersFinal equation would come out to be the one posted earlier..
bhai unable 2 get u completely.......(my inabilty) but still ur method has shown me a direction of approaching these kind of questions ab jab madad kr hi rhe ho toh.....yeh bhi dekh lo zara(Condition:No OA no options) Six persons try to swim across a wide river.Its known that on an average only three persons out of 10 are successful in crossing the river.Whats the probability that at most four of the 6 persons will cross the river safely?
bhai unable 2 get u completely.......(my inabilty)but still ur method has shown me a direction of approaching these kind of questionsab jab madad kr hi rhe ho toh.....yeh bhi dekh lo zara(Condition:No OA no options)Six persons try to swim across a wide river.Its known that on an average only three persons out of 10 are successful in crossing the river.Whats the probability that at most four of the 6 persons will cross the river safely?
Six persons try to swim across a wide river.Its known that on an average only three persons out of 10 are successful in crossing the river.Whats the probability that at most four of the 6 persons will cross the river safely?
Yeah.. Ans is coming out to be 0.98 ..Itna zada?? posssible?
Why not. If out of 10, only 3 are most probable to cross, hence in case of 6, the most of the favorable cases are there in P(1 cross), P(2 cross) [Note the power of 7 in each]. So, it will be highly likely that at most 4 cross.
Why not. If out of 10, only 3 are most probable to cross, hence in case of 6, the most of the favorable cases are there in P(1 cross), P(2 cross) [Note the power of 7 in each]. So, it will be highly likely that at most 4 cross.
Yeah..should be the right answer if you too got it 😛 I was just skeptical after getting such high probability
Yeah..should be the right answer if you too got it I was just skeptical after getting such high probability
Oh God, don't trust for calculations for me .. I have made mistakes as silly as 4*2 = 6 !
Find the value of C0 - C2 + C4 - C6 + C8................ using binomial theorem C0,C2........ represents nC0, nC2........ (No OA, so show approach, well, you know what, one who is confident should always show approach no matter what 😐 )
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