in lotto draw balls 1-50 mixed together. machine randomly selects numbers 13,11,7,27,41. is the 6th number drown
A. more like to be odd than even
b. more likely to be even than odd
c. equally likely
i could not decide, as long as there even and odd numbers, there is a possibility to draw any. but, since now there is 25 even and 20 odd numbers are in the machines, maybe even is more likely? (25 out of 45 and for odd 20 out of 45.. ) any insights?
originally there were 25 odds and 25 evens,
the 5 numbers already drawn are all odd,
so there remains:
25 evens and only 20 odds
so for the next draw:
prob(odd) = 20/45
prob(even) = 25/45
so what do you think?
To determine the likelihood of the 6th number being odd or even, we need to consider the composition of the remaining numbers in the machine.
If we assume that the first 5 numbers drawn do not affect the probability of the 6th number, then we can focus on the remaining 45 numbers in the machine. Out of these 45 numbers, we need to determine the number of odd and even numbers.
You mentioned that there are 25 even numbers (2, 4, 6, 8, ..., 50) and 20 odd numbers (1, 3, 5, 7, ..., 49) in the machine. Therefore, the probability of drawing an even number is 25/45 and the probability of drawing an odd number is 20/45.
Based on these probabilities, we can now evaluate the options provided:
A. More likely to be odd than even: If the probability of drawing an odd number (20/45) is higher than the probability of drawing an even number (25/45), then this statement would be correct. However, since the probability of drawing an even number is higher, option A is incorrect.
B. More likely to be even than odd: As explained above, the probability of drawing an even number (25/45) is indeed higher than the probability of drawing an odd number (20/45). Therefore, option B is correct.
C. Equally likely: The probabilities of drawing an odd number and an even number are not equal, so option C is incorrect.
In conclusion, based on the given information, the 6th number is more likely to be even than odd (option B).