A SPINNER WITH FOUR EQUAL SECTIONS NUMBERED 3,5,7 AND 8 IS SPUN TWICE. WHAT IS THE PRBABILITY THAT AN ODD NUMBER IS SPUN THE FIRST TIME AND AN EVEN NUMBER IS PUN THE SECOND? IS THE ANSWER 3/4? IS IT THAT EASY OR NOT?
nope. You have to account for both events.
P(odd) = 3/4
P(even) = 1/4
so, P(odd,even) = 3/4 * 1/4
To find the probability, we need to determine the total number of possible outcomes and the number of favorable outcomes.
The spinner has four equal sections numbered 3, 5, 7, and 8. Since there are four equally likely outcomes, the total number of possible outcomes is 4.
For the first spin, we want an odd number, which includes the numbers 3, 5, and 7. Therefore, the number of favorable outcomes for the first spin is 3.
For the second spin, we want an even number, which includes only the number 8. So the number of favorable outcomes for the second spin is 1.
To find the probability, we divide the number of favorable outcomes by the total number of possible outcomes:
Probability = Number of favorable outcomes / Total number of possible outcomes
Probability = (3 for first spin) * (1 for second spin) / (4 total)
Probability = 3/4
So yes, the answer is indeed 3/4. It is that simple in this case.