suppose you have 2 dimes ad 3 quarters in a bag. You shake it so that 2 coins fall out. 1) What is the probability that both coins are quarters? 2) both coins are dimes? 3) both coins are the same? 4)you shake out one of each coin?
20 things could happen in the two tosses
for example
D1 first
then
D2 or Q1 or Q2 or Q3
or
D2 first
then
D1, Q1 Q2 Q3
or
Q1 first
then
D1 D2 Q2 Q3
.
. etc
Both quarters out of the 20 I see
6 so 6/20 = 0.3
both dimes I see 2 out of the 20 or 0.1
both dimes or both quarters I see 8 out of the 20 so 0.4
one of each I see 12 out of the 20 or 0.6
How do you get 20.
To find the probabilities in this scenario, we need to understand the total number of possible outcomes and the number of favorable outcomes for each event.
Let's answer each question step-by-step:
1) What is the probability that both coins are quarters?
To determine the probability of both coins being quarters, we need to consider the total number of possible outcomes and the number of favorable outcomes.
Total number of possible outcomes: There are a total of 5 coins in the bag, so the total number of ways to choose 2 coins from 5 is given by the combination formula "5 choose 2," which is written as 5C2 and is calculated as (5!)/(2!(5-2)!), where "! "denotes factorial.
Favorable outcomes: Since we want both coins to be quarters, we only have 3 quarters to choose from, so the number of ways to choose 2 quarters from 3 is given by the combination formula "3 choose 2," which is 3C2 and is calculated as (3!)/(2!(3-2)!).
Probability of both coins being quarters: The probability is equal to the number of favorable outcomes divided by the total number of possible outcomes. So, the probability is 3C2 / 5C2.
2) What is the probability that both coins are dimes?
Following the same logic as above:
Total number of possible outcomes is still 5C2 since it is still a total of 5 coins.
Favorable outcomes: In this case, we only have 2 dimes to choose from, so the number of ways to choose 2 dimes from 2 is 2C2.
Probability of both coins being dimes: The probability is 2C2 / 5C2.
3) What is the probability that both coins are the same?
For both coins to be the same, we need either 2 quarters or 2 dimes.
Probability of both coins being quarters: We already calculated this in question 1, and the probability is 3C2 / 5C2.
Probability of both coins being dimes: We already calculated this in question 2, and the probability is 2C2 / 5C2.
We add the probabilities of both cases since they are mutually exclusive (they cannot occur at the same time):
Probability of both coins being the same: (3C2 / 5C2) + (2C2 / 5C2).
4) What is the probability that you shake out one of each coin?
To find this probability, we need to consider the total number of possible outcomes and the number of favorable outcomes.
Total number of possible outcomes: There are 5 coins, and we are selecting 2. So, the total number of possible ways to choose 2 coins from 5 is 5C2, as mentioned earlier.
Favorable outcomes: To shake out one of each coin, we need to select 1 quarter and 1 dime.
Number of ways to choose 1 quarter from 3 is 3C1.
Number of ways to choose 1 dime from 2 is 2C1.
To find the number of favorable outcomes, we multiply these two combinations since the events are independent (one doesn't affect the other).
Probability of shaking out one of each coin: (3C1 * 2C1) / 5C2.