Probability of Sequential Events
To solve this probability question, we must consider that the events are sequential and assume that, unless stated otherwise, a chosen number is not replaced. The first event is picking a 6 and the second event is picking a 5.
The probability of picking a 6 first:
There are 8 numbers in total. Only one of these numbers is a 6. So, the probability of picking a 6 is 1 out of 8, or:
\[ P(6) = \frac{1}{8} \]
Assuming the 6 is not replaced, we now have 7 numbers left.
The probability of picking a 5 next:
Since one number (the 6) has already been picked and is not being replaced, there are now 7 numbers remaining. Again, only one of these numbers is a 5. So, the probability of picking a 5 next is 1 out of 7, or:
\[ P(5) = \frac{1}{7} \]
To find the overall probability of both events happening in sequence (picking a 6 and then a 5), you multiply the probabilities of each individual event:
\[ P(6 \text{ and then } 5) = P(6) \times P(5) = \frac{1}{8} \times \frac{1}{7} \]
\[ P(6 \text{ and then } 5) = \frac{1}{56} \]
Therefore, the probability of picking a 6 and then picking a 5 is \( \frac{1}{56} \).
