There are 5 red shoes and 4 black shoes in a sale. They have got all mixed up with each other. What is the probability of getting a matched shoe if two shoes are drawn at random.
ans:4/9
(what i came up with was:
P(matched shoes)= 1- P(Not matched shoes)
= 1- { /9C2}
= 1- 5/9
Ans = 4/9)
Please let me know if i am correct and let me know your approach if i am not.
a person draw a card from a pack of 52, replaces it and shuffles it. he continue doing it untill he draws a heart. what is the probability that he has to make 3 trials.
ans: 9/64
please do post your approach.
Another one
Ram, Rahim, Guruteg, Zaru, are four speakers among 7 speakers at ZMSJ school of Business. Ram wants to speak before Guruteg, and Zaru wants to speak before Rahim. In how many orders can the speeches be scheduled?
a. 2520
b.5040
c. 1260
d. None of the above
Ram, Rahim, Guruteg, Zaru, are four speakers among 7 speakers at ZMSJ school of Business. Ram wants to speak before Guruteg, and Zaru wants to speak before Rahim. In how many orders can the speeches be scheduled?
a. 2520
b.5040
c. 1260
d. None of the above
Another one
Ram, Rahim, Guruteg, Zaru, are four speakers among 7 speakers at ZMSJ school of Business. Ram wants to speak before Guruteg, and Zaru wants to speak before Rahim. In how many orders can the speeches be scheduled?
a. 2520
b.5040
c. 1260
d. None of the above
Answer C 1260:drinking:
Another one
Ram, Rahim, Guruteg, Zaru, are four speakers among 7 speakers at ZMSJ school of Business. Ram wants to speak before Guruteg, and Zaru wants to speak before Rahim. In how many orders can the speeches be scheduled?
a. 2520
b.5040
c. 1260
d. None of the above
Is it c) 1260 ?
Reason:
First, let us select 4 places among the 7 places to organize these 4.
This can be done in 7C4 ways.
Now, considering the 4 places thus selected, we can choose any two places and arrange Ram and Guru. Ram and Guru can be organized in only one way, i.e Ram must come before Guru.
So, this can be done in 4C2 ways.
The rest of the two places can be occupied by Zaru and Rahim in only one way: Zaru coming before Rahim.
Thus, we can arrange these 4 in 7C4*4C2 ways.
The rest of the 3 participants can be arranged in 3! ways.
So, total number of ways: 7C4*4C2*3! = 1260.
P.S: Please correct me if I am wrong.
a person draw a card from a pack of 52, replaces it and shuffles it. he continue doing it untill he draws a heart. what is the probability that he has to make 3 trials.
ans: 9/64
please do post your approach.
He has to make 3 trials before drawing out a heart.
So, in the first two trials he must draw anything else but not the heart.
Probability of choosing any other card except the heart: 39/52 = 3/4
Probability of drawing a heart: 13/52 = 1/4
He has to draw any two cards except the heart for the first two.
So, prob = (3/4)*(3/4) = 9/16
He has to draw a heart in the third trial.
So, total prob = (9/16)*(1/4) = 9/64
Hope it helped..
a fair dice is tossed six times. find the probability of getting a third six on the sixth throw
ans: (5C2*5^3)/6^6
please do mention your approach
He has to obtain the thrid six in the last throw. So, he must obtain exactly 2 sixes in any of the first five throws.
We can select any 2 of the 5 throws by: 5C2.
These two trials have to get 6.
The probabiltiy of getting one six = 1/6.
So, prob of getting 2 sixes in these 2 trials = 1/6*1/6
Now, the rest of the three throws out of the first 3 throws must get any number other than 6.
Prob of getting any number other than 6 = 5/6.
So, getting any other number for 3 throws = 5/6*5/6*5/6.
Now, the last trial must get 6.
Prob = 1/6.
So, total prob = 5C2*(1/6)(1/6)(5/6)(5/6)(5/6)(1/6) = (5C2*5^3)/6^6
Hope it helped...
hi all,
can anyone explain me the concept of grouping and distribution and also the the formulas in them .
Thanks in advance
Another one
Ram, Rahim, Guruteg, Zaru, are four speakers among 7 speakers at ZMSJ school of Business. Ram wants to speak before Guruteg, and Zaru wants to speak before Rahim. In how many orders can the speeches be scheduled?
a. 2520
b.5040
c. 1260
d. None of the above
OA - C -1260
No of ways of arranging 7 speeches - 7! = 5040
OF THESE ways in how many ways is Ram always ahead of Guruteg - 1/2 * 5040
Of the remaining the number of ways in which Zaru speaks ahead of Rahim = 5040 * 1/2 *1/2 = 1260
Q1. A pair of fair dice are rolled together till a sum of either 5 or 7 is obtained. The probability that 5 comes before 7 is?
ans: 0.4
please do post your approach, thanks
the probability of a bomb hitting a bridge is 1/2 and two direct hits are needed to destroy it. the least number of bombs required so that the probability of the bridge being destroyed is greater than 0.9 is
ans: 7 bombs
please do post your approach.
Ques: A word of 6 letters are formed from a set of 16 different letters of the english alphabets(with replacements). find out the probability that exactly 2 letters are repeated.
ans: (225*224*156)/ 16^6
please do post your approach.
according to me ans should be . (6C2*16*16*15*14*13*12)/16^6. where as in the book answer, there should be only one 16 in the above derived answer. please help
Q1. A pair of fair dice are rolled together till a sum of either 5 or 7 is obtained. The probability that 5 comes before 7 is?
ans: 0.4
please do post your approach, thanks
Total number of ways in which 5 can come:
(1,4),(2,3),(3,2),(4,1) --- 4.
Total number of ways in which 7 can come:
(1,6),(2,5),(3,4),(4,3),(5,2),(6,1) --- 6.
So, total number of possible outcomes = 10.
The game stops when either one of them is obtained.
In this case, we need 5 before 7, which effectively means that we must obtain 5 because after that the game stops.
So, number of ways in which 5 can be obtained: 4.
Total number of ways = 10.
So, prob = 4/10 = 0.4
the probability of a bomb hitting a bridge is 1/2 and two direct hits are needed to destroy it. the least number of bombs required so that the probability of the bridge being destroyed is greater than 0.9 is
ans: 7 bombs
please do post your approach.
The prob of bomb hitting the bridge = 1/2.
2 bombs are required so that the bridge is destroyed.
The probability that if we fire 2 bombs and both hit the bridge is: 1/2*1/2 = 1/4.
So, in this case, if we stop bombing, the probability that the bridge is destroyed is only 25%.
So, if we fire another set of two bombs, and either of the set hits the bridge, the probability increases to:
1/4+1/4 = 1/2.
So, now there is a 50% chance of the bridge getting destroyed.
Again, if we fire a set of two bombs, the prob becomes: 3/4. (Total bombs = 6)
And, if we fire another set, the prob becomes: 1.(Total bombs =
So, it should be between 6 and 8, which is 7.
Ams employs 8 professors on their staff. Their respective probability of remaining in employment for 10 years are 0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9 .The probability that after 10 years atleast 6 of them still work in Ams is ???
a) 0.19 b) 1.22 c) 0.1 d) 0.16
please do post your approach.
hilips, the largest white goods producer in India, uses a quality check scheme on produced items before they are sent into the market. the plan is follows: A set of 20 articles is readied and 4 of them are chosen at random. if anyone of them is found to be defective then the whole set is put under 100% screaning again,if no defectives are found, the whole set is sent into the market. find the probability that a box containing 4 defective article will be sent into the market
ans: 364/969
PF/QA/ Set -1
An integer between 1 and 300, inclusive, is chosen at random. What is the probability that the integer so chosen equals an integer raised to an exponent that is an integer greater than 1?
(A) 17/300
(B) 1/15
(C) 2/25
(D) 1/10
(E) 3/25
An integer between 1 and 300, inclusive, is chosen at random. What is the probability that the integer so chosen equals an integer raised to an exponent that is an integer greater than 1?
(A) 17/300
(B) 1/15
(C) 2/25
(D) 1/10
(E) 3/25
Using digits 2,3,4,5 and 6 ,find the following.
a. Sum of all 5-digit numbers that can be formed such that no 2 digit are the same.
b. Sum of all 4-digit numbers that can be formed such that no 2 digit are the same.
c. Sum of all 4-digit numbers that can be formed such that digits can be repeated.
Please kindly answer to the above question with explanations.
We need find the number of integers in the form a^b where b is greater than 1.
Now nothing has been said about 'a' as far I understand so
So take all the squares 1,4,9.......289 ->17
Cubes (but cubes of a square cannot be taken sq^cube is a square and all squares are already there ) ->cubes of 2,3,5,6, ->4
Forth powers are taken into squares .
5th power 2^5 & 3^5
So total=17+4+2=23
And answer 23/300
Not in the options but I can't see anything wrong in the solution though !
Missed out 2^7. Tht makes total cases = 23+1
Prob = 24/300 = 2/25
Using digits 2,3,4,5 and 6 ,find the following.
a. Sum of all 5-digit numbers that can be formed such that no 2 digit are the same.
b. Sum of all 4-digit numbers that can be formed such that no 2 digit are the same.
c. Sum of all 4-digit numbers that can be formed such that digits can be repeated.
Please kindly answer to the above question with explanations.
Take a digit. Find out how many times it can be placed at a Unit's, Tens, Hundreds place... and so on postions uptil the no. of digits the resulting number is to comprise of. All other digits would also be placed same no. of times at each position. Add up the sum result and multiply with 111...
where for eg, for a particular 4 digit no. 1111 is nothing but representation of each position as 1111 = 1000 + 100 + 10 + 1
a) Sum = (2+3+4+5+6) * 4! * (11111)
b) Sum = (2+3+4+5+6) * C(4,3)*3! * (1111)
c) Sum = (2+3+4+5+6) * 4^3 * (1111)
will leave out further math..
PF/QA/ Set -1
An integer between 1 and 300, inclusive, is chosen at random. What is the probability that the integer so chosen equals an integer raised to an exponent that is an integer greater than 1?
(A) 17/300
(B) 1/15
(C) 2/25
(D) 1/10
(E) 3/25
Solution -
for squares all numbers from 1 to 17 are counted
for cubes - 2^3, 3^3, 5^3 and 6^3 are counted
For 4th power - none are added ( all of them will be considered as squares)
For 5th power -2^5, 3^5 are considered
For 6th power none of the numbers are considered
For 7th power only 2^ 7 is considered
Total favourable cases = 24
Ans = 24/300 which is equal to 2/25