A jar contains 4 red and 3 green jelly beans. Two jelly beans are taken simultaneously\What is the probability that both are the same color?

Question

A jar contains 4 red and 3 green jelly beans. Two jelly beans are taken simultaneously and completely at random from the jar. What is the probability that both are the same color?

Answer 1

Here is my approach. I believe that it is the easiest to understand. The student should answer the question in the way that makes the most sense.

 

There are 4 red candies and 3 green candies to choose from. No other candies are mentioned, so I will make the assumption that we are choosing just from these 7 candies.

 

We need to determine what is the probability of choosing 2 red ones. We also need to determine the probability of choosing 2 green ones. We then need to sum these two individual probabilities to get our answer.

 

2 red: P(1st one is red) = 4/7 followed by P(2nd one is red) = 3/6. The probability for this to occur is 12/42

2 green: P(1st one is green) = 3/7 followed by P(2nd one is green) = 2/6. The probability for this to occur is 6/42

 

Finally, the sum f these two is 18/42 which can be reduced to 3/7.

 

Thus, the probability that they will "match" is 3/7.

 

Therefore, the probability that you will get 2 distinct colors is 4/7.  (1 - 3/7) 

Answer 2

4 red 3 green

The total number of ways of choosing two is ₇C₂ = 21.

The number of ways of choosing 2 red out of 4 red is ₄C₂ = 6

The number of ways of choosing 2 green out of 3 green is ₃C₂ = 3

The number of ways of choosing two of the same color is 6 + 3 = 9

The probability is 9/21 = 3/7

 

4 red 3 green
The total number of ways of choosing two is ₇C₂ = 21.

The number of pairs where the beans are of opposite color is 4 × 3 = 12

(21 - 12)/21 = 9/21 = 3/7

 

Answer 3

Use of the word "simultaneously" indicates that we are to assume that neither jelly bean is chosen first, and therefore remaining choices do not decrease or increase in probability. I'm suspicious that this might not have been the intention, but I can only use the scenario as given.
There are two types of probability problems: "and" & "or". Your other one, involving a boy & a girl erasing a blackboard, is an "and" problem - as in "a girl & a boy". "And" problems are more exacting, specific, while "or" problems portray flexibility - whether we get "this one" or "that one" doesn't matter. Thus, "and" problems require multiplying fractions & yielding a smaller quantity than either of those fractions, while "or" problems involve addition, to increase the probability due to the options.
This one is an "or" problem. We want two jelly beans of the same color, but they can be "green or red". The probability of two red ones will be 4/7x4/7, there being 4 red jelly beans among a total of 7 jelly beans & we want 2 red ones. Likewise, the probability of choosing 2 green ones would be 3/7x3/7. Now we add the two products:
16/49 + 9/49 = 25/49
[Technically, it would be (4/7x3/7)+(3/7x2/7)=18/49, but the problem specified that we take the jelly beans "simultaneously" - it's a bit odd, but keeps the scenario free of more difficult concepts].
Check to see if simplification is necessary - in this case, it is as simple as it can be. So, the probability of getting 2 red or 2 green jelly beans in one draw, in the given scenario, is slightly more than 1/2.
Think of it this way: there's a slightly greater probability of getting "2 red or 2 green" than of getting exactly "1 red & 1 green". This might seem counter-intuitive at first glance, but remember there are fewer green jelly beans in the batch. The extra red jelly bean makes all the difference - it increases the probability of choosing 2 red ones & decreases the probability of getting a green one at all.