from 1-25 there are 25 numbers 1 3 5 7 9 11 13 15 17 19 21 23 25 = 13 numbers [2 4 6 8 10 12 14 16 18 20 22 24 = 12 numbers] so probability of selecting an odd number is 13 in 25 13:25
The probability is 21/36 = 7/12
It is 14/20 or 7/10.
Assuming he has to pick a whole number his possible choices are 5, 6, 7, 8, 9, 10 and 11. Four of these seven possibilities are odd numbers so the probability of him choosing an odd number is 4/7.
Let P(A) = 1/10; P(A) = probability of selecting one people on a basketball team P(B) = 1/35; P(B) = probability of selecting one people on a football team P(C) = 1/10 = probability of selecting one people who plays in both teams P(D) = probability of selecting from either team. P(D) = P(A) + P(B) - P(C) P(D) = 1/10 + 1/35 - 1/10 P(D) = 1/35 or 0.0286
It is 0.4
from 1-25 there are 25 numbers 1 3 5 7 9 11 13 15 17 19 21 23 25 = 13 numbers [2 4 6 8 10 12 14 16 18 20 22 24 = 12 numbers] so probability of selecting an odd number is 13 in 25 13:25
The probability is 21/36 = 7/12
It is 0.5
It is 14/20 or 7/10.
Assuming he has to pick a whole number his possible choices are 5, 6, 7, 8, 9, 10 and 11. Four of these seven possibilities are odd numbers so the probability of him choosing an odd number is 4/7.
of course... there is sometimes a odd number of 10's
The Probability of Success = Number of successful outcomes/Number of outcomes.E.g. Find the probability of choosing a red five or a black odd numbered card in a standard deck of 52 playing cards.There are 2 red fives and 10 odd numbered cards, and a total of 52 cards, so:=2/52+10/52=12/523/13Probability of Red 5 or a Black Odd = 3/13* * * * *The above is true only for discrete distributions, not for continuous variables. For a continuous variable, with probability distribution function p(x), the probability that x lies between two values, a and b (ie a
25/50 gives the probability of selecting a blue marble
The biggest odd number between 1 and 10 is 9.
Odd stupid
Let P(A) = 1/10; P(A) = probability of selecting one people on a basketball team P(B) = 1/35; P(B) = probability of selecting one people on a football team P(C) = 1/10 = probability of selecting one people who plays in both teams P(D) = probability of selecting from either team. P(D) = P(A) + P(B) - P(C) P(D) = 1/10 + 1/35 - 1/10 P(D) = 1/35 or 0.0286