Three cards are drawn from a well shuffled pack of 52 cards. Find the probability that all the three cards are of different suits and there is at least one face card.1) 1017/1105 2) 1197/5505 3) 129/1105 4) 1197/5525 5) 1017/5505
Total Number of ways in which the cards can be selected = 52C3 = 22100(Sample Space)
Nw, Number of ways in which 3 cards can be choosen following the face card requirement = 4C3*13^3 - 4C3*10^3 = 1197*4
Thus, P(E) = 1197*4/22100 = 1197/5505.. Par i am nt cnfdnt on this.. :(
Total Number of ways in which the cards can be selected = 52C3 = 22100(Sample Space)Nw, Number of ways in which 3 cards can be choosen following the face card requirement = 4C3*13^3 - 4C3*10^3 = 1197*4 Thus, P(E) = 1197*4/22100 = 1197/5505.. Par i am nt cnfdnt on this..
There are 13 cards in each suits n there are a total of 4 different suits in a pack of perfect cards..Thus, Each pack has 52 cards in it..
Nw, the question asks us to find the Probability of choosing 3 cards such that all the three belong to different suits n atleast one of the cards choosen is a face card[J Jack, K King, Q Queen]
Thus, Total Number of ways in which 3 cards can be choosen which belong to different 3 suits = 4C3(Ways of choosing 3 suits out of 4)*13^3(Ways on choosing 3 cards out of each suit)
Thus, Number of ways in which 3 cards can be choosen which belong to different 3 suits n have no face card= 4C3(Ways of choosing 3 suits)*10^3(Ways of choosing the other 10 cards barring the face cards)
Thus, Ways in which atleast 1 Face card is choosen= Total Number of ways - Number of ways in which none face card are choosen..
There are 13 cards in each suits n there are a total of 4 different suits in a pack of perfect cards..Thus, Each pack has 52 cards in it..Nw, the question asks us to find the Probability of choosing 3 cards such that all the three belong to different suits n atleast one of the cards choosen is a face card[J Jack, K King, Q Queen]Thus, Total Number of ways in which 3 cards can be choosen which belong to different 3 suits = 4C3(Ways of choosing 3 suits out of 4)*13^3(Ways on choosing 3 cards out of each suit)Thus, Number of ways in which 3 cards can be choosen which belong to different 3 suits n have no face card= 4C3(Ways of choosing 3 suits)*10^3(Ways of choosing the other 10 cards barring the face cards)Thus, Ways in which atleast 1 Face card is choosen= Total Number of ways - Number of ways in which none face card are choosen..Hence = 4*13^3 - 4*10^3 = 1197*4...
ohhh.......i was considering ace as a face card...
Probability of solving a question correct is 2/3. Also it is known that, if answer is not known then guess will be made. probability of guessing a question right is 1/4. so if we know that a question is answered & it is correctly answered. find the probability of him knowing the answer?
Probability of solving a question correct is 2/3. Also it is known that, if answer is not known then guess will be made. probability of guessing a question right is 1/4. so if we know that a question is answered & it is correctly answered. find the probability of him knowing the answer?
Copied from Testfunda :-The calendar in the cell phone of Mr.Donny has malfunctioned and he does not like paper calendars in his office nor does he like to see the dates on any other electronic equipment. His assistant sees Mr. Donny stuck in the quandary. To help him, he glances across Mr. Donny €˜s table and finds two different coloured dice but none of its faces have any numbers. He writes one number each on all the faces of both the dice so that Mr. Donny can set the dice each morning such that the top faces of the dice show the current date.What should be the numbers on each die to allow Mr.Donny to see any date?(Note: Single digits cannot be represented by just 1 die. On all days both dice must be used. So 9th would be 09).
Copied from Testfunda :-The calendar in the cell phone of Mr.Donny has malfunctioned and he does not like paper calendars in his office nor does he like to see the dates on any other electronic equipment. His assistant sees Mr. Donny stuck in the quandary. To help him, he glances across Mr. Donny €˜s table and finds two different coloured dice but none of its faces have any numbers. He writes one number each on all the faces of both the dice so that Mr. Donny can set the dice each morning such that the top faces of the dice show the current date.What should be the numbers on each die to allow Mr.Donny to see any date?(Note: Single digits cannot be represented by just 1 die. On all days both dice must be used. So 9th would be 09).
on one die the numbers should be(0,1,2,3,4,5) and on the other die numbers should be (1,2,6,7,8,9)
In a triangle ABC, a : b : c is 4 : 5 : 6 where a, b, c are the lengths of the sides opposite to angles A, B and C respectively then 3A + B is equal to ? PLEASE EXPLAIN THE METHOD !
A three-digit no when multiplied by 2, gives a four-digit no with at least one digit as 2. How many such three-digit nos are possible ??1) 176 2) 200 3) 576 4) 324 5) NOT
In a triangle ABC, a : b : c is 4 : 5 : 6 where a, b, c are the lengths of the sides opposite to angles A, B and C respectively then 3A + B is equal to ? PLEASE EXPLAIN THE METHOD !
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Ratio of sides 4:5:6 will be same as that of corresponding angles.
In a triangle ABC, a : b : c is 4 : 5 : 6 where a, b, c are the lengths of the sides opposite to angles A, B and C respectively then 3A + B is equal to ? PLEASE EXPLAIN THE METHOD !
hey guys,mine is a very basic question.i just wanna know the algorithm behind euclid's long divison method for finding hcf.i just want to know the logic behind this method of first dividing the bigger number by smallernumber(in case of two numbers) and then dividing the divisor by the coming remainder and so on.......y do we divide the divisor by the remainder(what's the logic behind this!?!) and how come we get hcf by this continuous subtraction and division?? (though hcf comes out by finding factors of the numbers!!!!! )p.s. i know its unnecessary but i just want to understand the concept in detail!! please explain...
First let us notice that if we take two co-prime numbers a and b (with a>b), then c = a – b is co-prime to both a and b.
Let's suppose that we are trying to find the HCF of two numbers. Let the required HCF be x. Then the two numbers can be written as a*x and b*x, where a and b are co-prime. Dividing the larger by the smaller gives a remainder (c)*x where c = (a – b) [Note: c could be (a-2b) or (a-3b) or whatever – the point is that this value is co-prime to a and b both!].
Now we could divide either of the original numbers by c*x [we usually choose the smaller number b*x for convenience] to get a new remainder d*x where d is also co-prime to b and c and so on. At some point we are going to divide two numbers and find a remainder zero, at this point the numbers must be k*x and x itself i.e. the last number is the HCF.
For example if we want HCF of 170 and 200 [note for reference that it is 10], 200 / 170 leaves 30, 170/30 leaves 20, 30/20 leaves 10 and 20/10 leaves 0. Note that at each step, the three numbers involved are co-prime multiples of 10.
We could also have used the large number; it wouldn't have mattered. In some cases, in fact, it could make it simpler (if one is alert) – for example to find the HCF of 1730 and 1557, 1730/1557 leaves 173….but 1730 is visibly 173*10 and so 1557 must be 173 *9 and hence the HCF must be 173.
A three-digit no when multiplied by 2, gives a four-digit no with at least one digit as 2. How many such three-digit nos are possible ??1) 176 2) 200 3) 576 4) 324 5) NOT
Please post the solution if you have any . I guess no one have solved till now.
Probability of solving a question correct is 2/3. Also it is known that, if answer is not known then guess will be made. probability of guessing a question right is 1/4. so if we know that a question is answered & it is correctly answered. find the probability of him knowing the answer?
