Last digit of 4^21 is?
At first you have to check the last number of 4&1, 4^2, 4^3, 4^4, 4^5. most of the last digit repeat in a series of 2 or 4. as you can see, the last digits are 4,6,4,6,4. That means the last digits is either 4 or 6. That means there are 2 possible last digits. Now divide 21 by 2. the remainder is 1. Above you can see that the last digit is 4 when remainder is 0 and last digit is 6 when remainder is 1. Therefore the answer is 6.
Proabaility that you get 3 heads when you flip a coin 5 times.
Either you can get a head or a tail. Probability of getting a head is 1/2 (p) and tail is 1/2 (1-p=q). Whenever you have a question that asks getting X (3 in above question) successful result out of Y (5 in above question) attempts with probability of success/failure, you have to solve using binomial theorem. The formula is: YCX * p^X * q^(Y-X) = 5C3 * (1/2)^3 * (1/2)^(5-3)
Find median: 9,5,4,7,6,3,3,9,4
Median is such a value that divides the set of numbers in two parts. Step 1: Sort the numbers in ascending order 3,3,4,4,5,6,7,9,9 Step 2: Count the numbers N=9 Step 3: the number 5 is in the middle of the numbers, i.e. four numbers (3,3,4,4) are on the left side of 5 and other four numbers (6,7,9,9) are in the right side of five. The formula is: median = ((N+1)/2)th term = (9+1)/2 th term = 5th term = 5