A party of n person sit at a round table. Find the odd against two specified person sitting next to each other??

If 4 whole numbers are taken at random and multiplied together, what is the chance that the last digit of the product is 1,3,7 or 9

In 4 throws with a pair of dice what is the chance of throwing a double twice?

A and B throw a pair of dice alternately. A wins if he throws 6 before B throws 5 and B wins if he throws 5 before A throws 6. Find B's chance of winning if A makes the first throw

2 person A and B toss a coin alternately till one gets head and win the game. Find probability of B's win if A tosses first?

10 pair of socks in a cupboard from which 4 individual socks are picked at random. The probability that there is atleast one pair is --

a fair coin is tossed 10 times. find the probability that 2 heads do not occur consecutively.

5 envelope and 5 letters placed at random. what is the probability that no single letter is placed in the right envelope

• n(E)=(n-2)!*2!
  n(s)=(n-1)! as circular arrngmnt
  nw
  P(E)=(n-2)!*2!/(n-1)!
  =2/(n-1)
  P(E)'=(1-(2/n-1))=(n-3)/(n-1)
  now ATQ
  odd against:P(E)'/P(E)=(n-3)/2
  ans=(n-3)/2
  
• 9 years agoHelpfull: Yes(18) No(0)
• 
  
  1)
  n(E)=(n-2)!*2!
  n(s)=(n-1)! as circular arrngmnt
  nw
  P(E)=(n-2)!*2!/(n-1)!
  =2/(n-1)
  P(E)'=(1-(2/n-1))=(n-3)/(n-1)
  now ATQ
  odd against:P(E)'/P(E)=(n-3)/2
  ans=(n-3)/2
  
  2)
  4 digit no's can be form in following ways=9*10*10*10
  only 1,3,7,9 can be multiplied to form 1,3,7,5 in end
  so favourable conditions = 4!
  p=(9*10*10*10-4!)/9*9*8*7
  
  3)
  double numbs are: 1,1 2,2 3,3 4,4 5,5 6,6. so total 6 types.
  now cases can be 4C2*(1*1/6 * 1*1/6 * 1*5/6 * 1*5/6) so ans is 25/216. rite?
  Alt. (6/36*6/36)
  
  4)
  chance of winning of b is.....5/6*1/6, 5/6*5/6*5/6*1/6.....
  p= 5/6*1/6 + 5/6*5/6*5/6*1/6 + 5/6*5/6*5/6*5/6*5/6*1/6 +..............
  p=1/6*5/6 [1+(5/6)^2+(5/6)^4+(5/6)^6.........)
  p=5/11
  
  
  5)
  
  chance of winning of b is..... 1/2*1/2 + 1/2*1/2*1/2*1/2 +...........
  p=1/3
  
  6)
  p'=(20C1*18C1*16C1*14C1)/(20C1*19C1*18C1*17C1)
  P=1-P'
  P=1-((20*18*16*14)/(20*19*18*17))
  
  7)2*(5c1(1/2)*4c1(1/2)*3c1(1/2)*2c1(1/2)*1c1(1/2))
  
  8)p=(4/5)^5
• 9 years agoHelpfull: Yes(12) No(1)
• qusn no-6
  
  
  take socks pair like that
  a b.....................j
  A B.....................J{total 20indivisual socks}
  take four indivisual socks out of 20
  =20C4.
  now
  
  n(E)=(20C1*18C1*16C1*14C1)/4!{we are focusing on differnt socks i.e unique comnitation and after 20 just bcz we cant take sam sacks pair like if we have a then can't take A. }
  P(E)=(20C1*18C1*16C1*14C1)/4!)/20C4
  =99/323
  ans:99/323
• 9 years agoHelpfull: Yes(7) No(1)
• Ans 8 :
  If we arrange the possibilities:
  
  1. Tosses 1,3,5,7,9 are tails
  The others can be either heads or tails: 2^5 possibilities.
  
  2. Tosses 2,4,6,8,10 are tails
  The others can be either heads or tails: 2^5 possibilities.
  
  Hence, total possibilites where no 2 consecutive are heads is: 2 * 2^5
  
  Probability = 2^6/2^10 = 1/2^4
  
• 9 years agoHelpfull: Yes(6) No(0)
• qno-8
  if we can put all letter correctly into the envelope into
  one way so and their problity
  =1/5!
  now
  no single letter is placed in the right envelope =1-1/5!=119/120
  ans=119/120
  
  
• 9 years agoHelpfull: Yes(4) No(4)
• answer of question 2 will be....... (0.4)^4..
  answer of question 4 will be......16/61 ....it will be infinite series.in gp.. a=16/81 and r will be...20/81 ...
  answer of question..5 will be 1/3..
• 9 years agoHelpfull: Yes(2) No(0)
• for question 1.
  total no of ways = (n-1)!
  when 2 specific person sits next to each other => (n-2)!*2!
  so odd against two = (n-3)/2
• 9 years agoHelpfull: Yes(2) No(0)
• sorry anser of question 4 wil be........5/11
  solution......A chance of winning is 1/6 and b chance of winning is 1/6...
  so the chance of winning of b is.....5/6*1/6, 5/6*5/6*5/6*1/6......simly..it will go to infinite..if we observe above pattern , we will find that it Is in gp..with first term=5/36 and r eqls to25/36...formula for sum of infinite series is a/1-r ....after putting the value in above equation we will get..5/11 .
• 9 years agoHelpfull: Yes(1) No(6)
• 3 one: double numbs are: 1,1 2,2 3,3 4,4 5,5 6,6. so total 6 types.
  now cases can be (1/6 * 5/6 *1/6* 5/6)4!/2!*2! so ans is 25/216. rite?
• 9 years agoHelpfull: Yes(1) No(0)
• what is the ans of second last questio.. ,ie..7 no
• 9 years agoHelpfull: Yes(1) No(0)
• 4.probability of getting double with pair of dice= 1/6
  probability of not getting double=5/6
  In total of throw we need double twice
  so, probaility=4C2 *(1/6)^2 *(5/6)^2
• 9 years agoHelpfull: Yes(1) No(0)
• 1) n-3/2
  2) 16/625
  3) 25/216
  4) 5/11
  5) 1/3
  6) 99/323
  7) 1/2^3
  8) 11/30
• 8 years agoHelpfull: Yes(1) No(0)
• For Question 2.
  
  There are 10 digits 0,1,2,........9 any of which can occur in any number at the last place.
  It is obvious that if the last digit in any of the four numbers is 0,2,4,5,6,8 then product of any such four numbers will not give a number having it's last digit as 1,3,7,9. Hence, it is necessary that the last digit in each of the four numbers must be any of the four digits 1,3,7,9.
  Thus, for each of the four numbers,
  the number of ways for the last digit = 10
  favourable number of ways = 4
  therefore, the probability that the last digit is any of the four numbers 1,3,7,9 = 4/10 = 2/5
  Hence, the required probability that the last digit in each of the four numbers is 1,3,7,9 so that the last digit in their product is 1,3,7,9 = (2/5)4 = 16/625.
• 6 years agoHelpfull: Yes(1) No(0)
• ans of first (n-2)!*2!
• 9 years agoHelpfull: Yes(0) No(1)
• ans of question 3 is 5/12
  chance of throwing a double-(11,22)...till(11,66)=5 cases
  -(22,33)...till(22,66)=4 cases
  -(33,44).....(33,66) =3 cases
  ans so on
  so total cases=15
  so prob=15/36= 5/12
• 9 years agoHelpfull: Yes(0) No(4)
• 2)
  4 digit no's can be form in following ways=9*10*10*10
  only 1,3,7,9 can be multiplied to form 1,3,7,5 in end
  so favourable conditions = 4!
  p=(9*10*10*10-4!)/9*9*8*7
• 6 years agoHelpfull: Yes(0) No(0)
• que. 2
  we know that 4 digit nos from 1000 to 9999 are=10000
  and only if we pick the nos which have combination of this digits 1,3,7,9 give us last digit after product
  So we can form 256 nos using 1,3,7,9 because repetition is allowed
  so 256/10000 =16/625
• 6 years agoHelpfull: Yes(0) No(0)
• a fair coin is tossed 10 times. find the probability that 2 heads do not occur consecutively.
  Ans --979/1024.
  In this que. we tossed coin for 10 times so total favourable cases for head/tail are =1/2^10= 1/1024.
  but 2 heads not occur consecutively so lets we find consecutive heads first
  for 2 consecutive heads=9 possible case.
  same for 3 consecutive heads=8
  for 4 consecutive heads=7.
  for 5 consecutive heads=6
  for 6 consecutive heads=5.
  for 7 consecutive heads=4.
  for 8 consecutive heads=3.
  for 9 consecutive heads=2.
  for 10 consecutive heads=1
  total favourable cases is 45
  so 45/1024 is favour of 2 or more consecutive heads
  
  1-45/1024 =979/1024 ans.
• 6 years agoHelpfull: Yes(0) No(0)