binish wrote: bhavin wrote: marcel wrote: i have 1 more query...
In how many ways can one choose 6 cards from a normal deck of cards so as to have all suits present?
a. (13^4) x 48 x 47 b. (13^4) x 27 x 47 c. 48C6 d. 13^4 e. (13^4) x 48C6
pls help...
Take a look at this
Bhavin,
I have a doubt in this question.In the selection of last 2 cards u r selecting in individual ways ie 48 & 47 ways........whats the error if we choose last 2 cards from 48 cards so selection would be 48C2......
soory for chipping in late guys...once everything is streamlined in a week or so...i shall be more regular ..
excellent binish...i was wanting for somebody to correct the approach long back...i had answered this question then, considering the ans option....actually none of the ans options are correct...and u have corrected the mistake also partially...
yes...your point is valid...if we have to make any 2 selections of the last 48 cards it has to be 48C2 and not 48*47..because which card u select first does not matter... But, 13^4 * 48C2 is also not correct !!
Refer back to the question u had raised some time back ...it read :
A box contain 5 Red & 6 White balls.In how many ways can 6 balls be selected so that there are atleast 2 balls of each color?
And i had said it is not 5C2 * 6C2 * 7C2 because of multiple selections...because 5th and 6th choice of balls were already accounted for when we considered the individual colors..
In similar way, 13^ 4 * 48C2 is incorrect due to multiple countings...
Let us list the no of ways restriction is met : 1) 3 cards of 1 suit and 1 card from remaining 3 suites ...
OR
2) 2 cards from 2 suites and 1 card from remaining 2 suites ..
These are the only 2 possibilities so that we have atleast 1 card from each suite..Let us count the no of ways each of the 2 conditions can be met...
1) we can select 1 out of 4 suites in 4C1 ways and 3 cards from that suite in 13C3 ways and then 1 card from remaining 3 suites in 13C1 ways each..
bhavin wrote: binish wrote: bhavin wrote: marcel wrote: i have 1 more query...
In how many ways can one choose 6 cards from a normal deck of cards so as to have all suits present?
a. (13^4) x 48 x 47 b. (13^4) x 27 x 47 c. 48C6 d. 13^4 e. (13^4) x 48C6
pls help...
Take a look at this
Bhavin,
I have a doubt in this question.In the selection of last 2 cards u r selecting in individual ways ie 48 & 47 ways........whats the error if we choose last 2 cards from 48 cards so selection would be 48C2......
soory for chipping in late guys...once everything is streamlined in a week or so...i shall be more regular ..
excellent binish...i was wanting for somebody to correct the approach long back...i had answered this question then, considering the ans option....actually none of the ans options are correct...and u have corrected the mistake also partially...
yes...your point is valid...if we have to make any 2 selections of the last 48 cards it has to be 48C2 and not 48*47..because which card u select first does not matter... But, 13^4 * 48C2 is also not correct !!
Refer back to the question u had raised some time back ...it read :
A box contain 5 Red & 6 White balls.In how many ways can 6 balls be selected so that there are atleast 2 balls of each color?
And i had said it is not 5C2 * 6C2 * 7C2 because of multiple selections...because 5th and 6th choice of balls were already accounted for when we considered the individual colors..
In similar way, 13^ 4 * 48C2 is incorrect due to multiple countings...
Let us list the no of ways restriction is met : 1) 3 cards of 1 suit and 1 card from remaining 3 suites ...
OR
2) 2 cards from 2 suites and 1 card from remaining 2 suites ..
These are the only 2 possibilities so that we have atleast 1 card from each suite..Let us count the no of ways each of the 2 conditions can be met...
1) we can select 1 out of 4 suites in 4C1 ways and 3 cards from that suite in 13C3 ways and then 1 card from remaining 3 suites in 13C1 ways each..
P.S : Vijay, i have copied your harder probability questions here :
Everybody, pls post all your probability related queries here !!
Here are the questions from Vijay :
A 1-inch-diameter coin is throw n on a table covered wit h a grid of lines tw o inches apart. What is the probabilit y t he coin lands in a square wit hout touching any of the lines of t he grid?
1/4 1/2 1/3 1/6 1/8
An airplane is built to be able to f ly on one engine. If the plane' s two engines operate independently, and each has a 1% chance of failing in any given f our-hour flight , w hat is the chance t he plane w ill fail t o complete a four-hour flight to Oklahoma due t o engine f ailure?
.0001 .001 .00001 .01 1
In a roomful of 30 people, what is the probabilit y t hat at least two people have the same birt hday? Assume birt hdays are uniformly dist ributed and there is no leap year complication.
70 60 50 35 80
How many 7-digit numbers have no two adjacent digits equal?
4,782,969 5,782,969 4,600,969 4,782,600 4,782,999
How many 5-digit palindromes are there? (A palindrome is a number that reads the same way forwards and backwards. For example, 27872 and 48484 are palindromes, but 28389 and 12541 are not.)
In a roomful of 30 people, what is the probabilit y t hat at least two people have the same birt hday? Assume birt hdays are uniformly dist ributed and there is no leap year complication.
70 60 50 35 80
Vijay, are the answer options correct ? Or are they supposed to be decimal nos ? Integer answer options for probability question is invalid ..Pls correct the answer option and repost this question ..
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