Prob(Pick 6 from 1-15 in 99 trials) = 1 - Prob(Don't pick 6 from 1-15 in 99 trials) = 1 - (14/15)99 = 1 - (0.933...)99 = 1 - 0.00108 approx = 0.9989
well it Will be even
1/6Because:There is a 4 in 9 chance of picking a red ball the first time and 3 in 8 chance of picking a red ball the second time. The chance of picking two reds as the first two balls is 4/9 x 3/8 or 12/72 or 1/6
The probability, in one draw, is 6/20 or 30%
The event space comprises the numbers 10 to 99, 90 such numbers. The favourable events are 15, 21, 27, ... , 99. There are 15 such numbers. So the probability is 15/90 = 1/6
Prob(Pick 6 from 1-15 in 99 trials) = 1 - Prob(Don't pick 6 from 1-15 in 99 trials) = 1 - (14/15)99 = 1 - (0.933...)99 = 1 - 0.00108 approx = 0.9989
well it Will be even
3 out of 13 or 3/13
The probability of any one number on a die being rolled is 1/6 or 16.67%.
There are two black 6's in a deck of cards, the probability of not picking one is 50/52 or 25/26, or 96.2%
11 marbles total and 6 are blue so probability is 6/11
It is 1/13,983,816.
0.4
1/6Because:There is a 4 in 9 chance of picking a red ball the first time and 3 in 8 chance of picking a red ball the second time. The chance of picking two reds as the first two balls is 4/9 x 3/8 or 12/72 or 1/6
3/5 probability of a white ball being drawn.
This is a law of addition probability which states that the probability of A or B equals the probability of A plus the probability of B minus the probability of A and B. Written in mathematical terms, the equation is: P(AorB) = P(A) + P(B) - P(AnB) where P(AnB) = 0 (since you can not pull out a green and black ball at the same time). Let P(A) = Probability of drawing the green ball & let P(B) = Probability of drawing the black ball. Total outcomes is 17. So, P(A) = 4/17 & P(B) = 6/17. Therefore P(green or black) = 4/17 + 6/17 = 10/17.
If the balls are selected at random, then the probability is 12/95.