Drawing a red marble from a bag of 6 red and 4 blue marbles, replacing it and then drawing a blue marble
count all the marbles together you start with so that's 10. 5/10 for blue marbles and then 5/10 for red
To solve this problem, we need to determine the probability of drawing a red marble first, then replacing it, and finally drawing a blue marble.
Step 1: Calculate the probability of drawing a red marble.
The bag contains a total of 6 red marbles and 4 blue marbles, so the probability of drawing a red marble on the first draw is:
P(Red on first draw) = (Number of Red Marbles) / (Total Number of Marbles)
= 6 / 10
= 0.6 or 60%
Step 2: Replace the first drawn marble back into the bag.
Since we are replacing the first drawn marble back into the bag, the total number of marbles remains the same.
Step 3: Calculate the probability of drawing a blue marble after replacing the first marble.
After replacing the first drawn red marble back into the bag, we now have 6 red marbles and 4 blue marbles. So the probability of drawing a blue marble on the second draw is:
P(Blue on second draw) = (Number of Blue Marbles) / (Total Number of Marbles)
= 4 / 10
= 0.4 or 40%
Step 4: Calculate the final probability of drawing a red marble and then a blue marble.
Since we are drawing marbles in two separate steps (first red, then blue) and both are independent events, we can multiply the probabilities from Step 1 and Step 3 to get the final probability:
P(Red and then Blue) = P(Red on first draw) * P(Blue on second draw)
= 0.6 * 0.4
= 0.24 or 24%
So, the probability of drawing a red marble from the bag, replacing it, and then drawing a blue marble is 0.24 or 24%.
I assume you want the probability of doing that...
6/10 * 4/10 = 24/100