It is equally probable That the pointer on the spinner shown will land on any one of the 8 regions number 1 through 8 if the pointer lands on a borderline spin again find the probability that the pointer will stop on an odd number or a number less than 5 answer with fractions simplify
There are 4 odd numbers (1, 3, 5, 7) and 4 numbers less than 5 (1, 2, 3, 4).
Out of the 8 regions, there are 2 regions (1 and 8) that land on an odd number, and 4 regions (1, 2, 3, 4) that land on a number less than 5.
Since landing on a borderline means to spin again, the probability is equal to the probability of landing on an odd number or a number less than 5 on the first spin, plus the probability of landing on a borderline and then getting an odd number or a number less than 5 on the second spin.
The probability of landing on an odd number or a number less than 5 on the first spin is:
2/8 + 4/8 = 6/8 = 3/4
Since any spin has an equal probability of landing on any of the 8 regions, the probability of landing on a borderline on the first spin is 1/8.
The probability of getting an odd number or a number less than 5 on the second spin, given that the first spin landed on a borderline, is:
(4/8) + (4/8) = 8/8 = 1
Therefore, the probability of landing on an odd number or a number less than 5, given that the pointer lands on a borderline, is:
1/8 * 1 = 1/8
So the final answer is 1/8.