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Algebra
Q1. Shreyas visits the cities A, B, C and D at random. What’s d probability that he visits
1) A before B
2) A just before B
Q2. A pack of 52 cards were distributed equally among 4 players.
Find the chance that 4 kings are held by a particular player.
Q3. 2 cards are drawn from a pack and kept out. Then 1 card is drawn from remaining 50 cards.
Find the probability that it is an ace.
Q4. Six dice are thrown 729 times. How many times do you expect at least 3 dice to show 5 or 6 ?
Comments (13)
solution--(2)
we know that 13 cards to be delivered.
total no. ways......52C13 =52! / (13! * 39!)
since 4 kings are to be held by a specified player,9 more cards are to be given to him
from remaining 48 cards 48C9 ways.
P=48C9 / 52C13
ans:-11/4165
here are the solutions
Q1 1 part
since shreyas want to visit A before B but not necessarily just before B
this statement means that A and B does want to arrange themsleves , their order must remain same that A before B
and the things which donot want to arrange can be taken as identical
in this Q as if A and B are identical and donot want to arrange themselves.
hence favourable cases = 4 ! / 2! ( two identical A and B )
total cases = 4! = 24
hence probability = 12/24 = 1/2
2 part
now in this case shreyas want to visit city A just before B means A and B should be together in order A before B
hence they are treated as one object
therefore favourable cases = 3! ( notice that 3! is not multiplied by 2! coz A and B does not want to change their order )
total cases = 24
hence probability = 6/24
answer to 3 Q
firstly two cards are drawn then probability of drawing an ace will depend upon what are those first tow cards
if first two cards are ace then we have only two ace remaining
if one is ace other is non ace then we hace only one ace remaining
if both are non ace then we have 4 aces reamaining
hence probability =
( probablity of both ace) ( probablity of one ace from reamining cards) + ( probability of one ace and one non ace) ( probability of one ace from reamaining) + (probablity of both non ace) ( probability of ace from remaining)
= [( 4C2 / 52C2 )* (2C1*/ 50C1) ] + [ (4C1*48C1/52C2) *( 3C1*50C1)] + [(48C2/52C2) * ( 4C1 *50C1)]
solving this u will get ans = 1/13 ..
Let P = probability (showing 5 or 6) = 2/6 = 1/3 q = 1 - p = 1- 1/3 = 2/3 n = 6 and r = 3 Also P (X= R) = probability (at least 3 dice will show 5 or 6 in one trial) Using the 'complement' theorem P(atleast three times)= 1- [P(X=0)+P(X=1)+P(X=2)] p (X = R) = 1 - [P (X = 0) + P(X = 1) + P (X = 2)]
I have a major doubt.....In the question 1):1
If he visits A first, then he can visit B second third or last.....if A is second,B is 3rd or 4th...If A is 3rd, B is last.....If A is last...No chance fr statment to be true......So total number of favourable situations is 3+2+1=6
All cases are simply 4!=24
So probability is 6/24=1/4 right??
Which cases am I missing??
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