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How do you randomly select 3 people out of 4?
ans- 4c3= 4!\3!*1! =4*3*2*1\3*2*1 =4
What is the probability of rolling 3 ones with 4 dice?
The probability of rolling 3 ones with 4 dice is:4C3 (1/6)3 (5/6) = 0.015432098... ≈ 1.54%
How many 3 digit numbers can make out of this 4 digit number 8426?
4C3 = (4!)/(4 -3)!3!] = (4 x 3!)/(1!3!) = 4 ways
A fair coin is tossed four times What is the probability that exactly three tosses are heads?
n = number of trials = 4 r = number of success = 3 P(success) = P(H) = p = 1/2 (in 1 toss of a fair coin) P(failure) = P(T) = q = 1/2 Use the formula nCr [p^(n-r)](q^r) : (4C3)[(1/2)^1](1/2)^3 = (4C3)(1/2)^4 = 4 x 1/16 = 4/16 = 1/4
What is the probability of getting heads on all 3 times if flips coin 4 times?
If a coin is flipped 4 times, the probability of getting 3 heads is: 4C3 (1/2)^3 (1/2)^1 = 4(1/8)(1/2) = 4/16 = 1/4
There are 8 railway stations along a railway line how many ways can a train be stopped at 3 of these stations such that no 2 of them are consecutive?
choosing 3 stations from 8... balance 8-3=5stations 5 stations have 4 gaps in between arrange these 3 (where it stops) in 4 gaps in 4C3=4 ways
What is the probability of a coin tossed four times to get 3 tails?
Using a calculator, plug numbers into the formula: nCk/2^n 4C3/2^4 = 4/16 = 1/4 Pascal's triangle can be used, if you are familiar with that (makes things much easier for a .5 probability scenario. See related link
What is the probability of getting one head and 3 tails on 4 coins?
My best thing Is to use a dime , that's the best luck you can get ... TRUST me , there Is NO Doubt (: & Throw the dime up in the air 4 times ------------------------------------------------------------------------------------------------ From Rafaelrz. The probability of getting one head and 3 tails on a 4 coin toss is; 4C3 x (1/2)4 = 4!/[3!(4-3)!] x [1/2]4 = 0.25 or 25%
What is the probability of having 3 boys for 4 kids?
Assuming probability of having a boy is P(B) = 1/2, and of having a girl is P(G) = 1/2,the probability of having 3 boys for 4 kids (with out regard to the girl to be the first,second, third or fourth kid) is;P(3B1G) = 4C3 [P(B)]4 = 4 (1/2)4 = 0.250 = 25%The factor 4 comes because there are 4 possibilities for the order in which the girl cancome out.
In how many ways can a 3 person committee be selected from four distinct pairs of husband and wives so as NOT to have spouse in it?
let the pairs be named as A1 B1,A2 B2,A3 B3,A4 B4.LET US SELECT TWO FROM BOYS AND ONE FROM GIRL THIS WE CAN DO IN 4c2*2c1=12 NOW THE OTHER WAY i.e 2 GIRLS AND 1 BOY 4c2*2c1=12 OTHER WAY IS TO SELECT ALL THE THREE FROM GIRLS OR BOYS=2*4c3 TOTAL 30
1 A person has 4 coins each of different denomination What is the number of different sums of money the person can form using one or more coins at a time?
Number of different sums of money that can person form is 15. Step-by-step explanation: Given : A person has 4 coins each of different denomination. To find : What is the number of different sums of money the person can form? Solution : A person has 4 coins each of different denomination. Selecting 1 coin out of 4 coins gives possible ways. Selecting 2 coin out of 4 coins gives possible ways. Selecting 3 coin out of 4 coins gives possible ways. Selecting 4 coin out of 4 coins gives possible ways. Total number of different sums of money that can person form is 4c1+4c2+4c3+4c4= 4+6+4+1=15 Therefore, Number of different sums of money that can person form is 15
Is the cube root of 2 irrational?
The cube root of 2 is irrational. The proof that the square root of 2 is irrational may be used, with only slight modification. Assume the cube root of 2 is rational. Then, it may be written as a/b, where a and b are integers with no common factors. (This is possible for all nonzero rational numbers). Since a/b is the cube root of 2, its cube must equal 2. That is, (a/b)3 = 2 a3/b3 = 2 a3 = 2b3. The right side is even, so the left side must be even also, that is, a3 is even. Since a3 is even, a is also even (because the cube of an odd number is always odd). Since a is even, there exists an integer c such that a = 2c. Now, (2c)3 = 2b3 8c3 = 2b3 4c3 = b3. The left side is now even, so the right side must be even too. The product of two odd numbers is always odd, so b3 cannot be odd; it must be even. Therefore b is even as well. Since a and b are both even, the fraction a/b is not in lowest terms, thus contradicting our initial assumption. Since the initial assumption cannot have been true, it must be false, and the cube root of 2 is irrational.
