A quarter, nickel, and dime were tossed. Above are the possible outcomes. What is the possibility of getting at least two heads?
1/2
7/8
3/8
5/8
In a bag, there are 10 black plastic chips, 7 yellow, 5 red, and 3 green. Two chips are drawn at random with replacement.
Find P(black, then red).
3/5
2/25
2/3
2/5
I don't know what you list consists of, but there should have been 8 of them
Count how many have exactly 2 H and include the 3 H
divide that sum by 8
for the 2nd, since you are returning the ball after the first draw, and simply
note the result, the denominator for both draws is 25
multiply the two probabilities and simplify the fraction
To find the possibility of getting at least two heads when tossing a quarter, nickel, and dime, we need to consider all the possible outcomes.
There are 2 possible outcomes for each coin (heads or tails), and since we have three coins, there are 2^3 = 8 possible outcomes in total.
Out of these 8 outcomes, we need to find the ones where at least two coins land on heads.
The possible outcomes that satisfy this condition are: HHT, HTH, THH, and HHH.
Therefore, the possibility of getting at least two heads is 4 out of 8, or 4/8, which simplifies to 1/2.
Answer: 1/2
To find the probability of drawing a black chip, then a red chip from a bag with 10 black, 7 yellow, 5 red, and 3 green chips with replacement, we need to consider the total number of chips and the number of black and red chips.
There are a total of 10 + 7 + 5 + 3 = 25 chips in the bag.
The probability of drawing a black chip is 10/25, as there are 10 black chips. After drawing a black chip and replacing it back in the bag, the number of chips remains the same.
The probability of drawing a red chip after drawing a black chip is 5/25, as there are 5 red chips left in the bag.
To find the probability of both events happening (drawing a black chip, then a red chip), we multiply the probabilities together.
P(black, then red) = (10/25) * (5/25) = 50/625 = 2/25
Answer: 2/25
To find the possibility of getting at least two heads when tossing a quarter, nickel, and dime, we need to determine the probability of each outcome and add them up.
There are three coins, and each coin has two possible outcomes: heads or tails.
So, the total number of possible outcomes is 2 * 2 * 2 = 8.
Now, let's examine the possible outcomes that result in at least two heads:
- HHT: Quarter shows heads, nickel shows heads, and dime shows tails.
- HTH: Quarter shows heads, nickel shows tails, and dime shows heads.
- THH: Quarter shows tails, nickel shows heads, and dime shows heads.
- HHH: Quarter shows heads, nickel shows heads, and dime shows heads.
Since there are four possible outcomes with at least two heads, the probability of getting at least two heads is 4/8, which simplifies to 1/2.
Therefore, the answer to the first question is 1/2.
Now, let's move on to the second question.
In a bag, there are 10 black plastic chips, 7 yellow, 5 red, and 3 green. Two chips are drawn at random with replacement, meaning after each draw, the chip is put back into the bag.
The probability of two independent events occurring is found by multiplying their individual probabilities.
First, let's find the probability of drawing a black chip: There are a total of 10 + 7 + 5 + 3 = 25 chips in the bag, and 10 of them are black. So, the probability of drawing a black chip on the first draw is 10/25.
Since the chip is put back into the bag after the first draw, the probability of drawing a red chip on the second draw remains the same: 5/25.
To find the probability of both events happening, we multiply the probabilities: (10/25) * (5/25) = 50/625, which simplifies to 2/25.
Therefore, the answer to the second question is 2/25.
Please let me know if you need any further clarification or explanation.