IIT JEE , AIEEE Forum - Mathematics , Algebra

sharmadon

1. there are 6 diff pairs of gloves.if 8 gloves are selected at random then what is the probability that there are exactly 2 pairs??

2. 12 distinct balls are distributed among 3 different boxes.then what is the probability that first box contains 3 balls?

3. a six-faced die is biased such that it is twice to show an even no. when thrown. if it is thrown thrice then what is the probability that sum of the no. is even?

4.there is a group of 6 persons.they play a game in which each has to select a number from 1 to 4.the group wins if at least 5 of them have same selection of numbers, then what is the probability that the group wins?

5.what is the probability of selecting two squares of 1*1 house from a chess board with one side in common?

Werewolf

This is your second question.

Each ball has three different ways to go to a box , to box A or B or C , so total no. of ways are 3¹²

Now we have to find the favorable events , choose 3 balls in ¹²C₃ ways and give it to box A , now remaining 9 balls have only two choices box B or C so they can be filled in 2⁹ ways so total fav ways are ¹²C₃*2⁹

Probability ¹²C₃*2⁹/3¹²

Werewolf

Let the probability of throwing and even number be 2x and of odd be x , then

2x+2x+2x+x+x+x =1 => x=1/9

Now in 3 trials we can get even sum if we throw odd+odd+even or even +even+even
So the probability is 3x*3x*6x+6x*6x*6x = 10/27

Werewolf

Answer to your 5th question is 1/18 , for solution u can refer this link

Click here

Werewolf

This is your fourth question

Probability of selecting 1 =prob of selecting 2=prob of selecting 3 =prob of selecting 4=1/4

Probability of atleast 5 people selecting 1 = prob of 5 people + prob of 6 people= (1/4)⁵(3/4)+(1/4)⁶
But since there are 4 such numbers so probability is to be multiplied by 4

Therefore final answer is 4[(1/4)⁵(3/4)+(1/4)⁶ ]

gaurav3sharma

the answer to the 4th question is..

total no. of ways 4^6

then the fav. cases...

for at least 5 =prob of 5 people + prob of 6 people

prob. of 5 people is no. of ways of choosin 5 people=6C5

then there will be three cases when 5 have same no. and 6th has the other no. thus for 4 nos. its 4*3

6C5 ( 4 * 3 ) ..

prob for 6 is only 4 cases..

hence reqd prob. is..

6C5*12 + 4 / 4^6

gaurav3sharma

is the answer to the first question is 12/33?

Werewolf

Post your solution

sharmadon

please help me with the 1st question...

Ask & Discuss Questions with Community & Experts