Without looking Jenny randomly selects two socks from a drawer containing four blue, three white, and five black socks, none of which are paired up. what is the probability that she chooses two socks of the same color?

Answer 1

#19/66#

We can add the probability of choosing two blue, two white or two black socks, since they are mutually exclusive possible outcomes.

The total number of socks is #4+3+5=12#

The probability of choosing two blue socks is:

#4/12*3/11 = 1/11#

The probability of choosing two white socks is:

#3/12*2/11 = 1/22#

The probability of choosing two black socks is:

#5/12*4/11 = 5/33#
In each case, the probability that the first sock is a particular colour is the number of socks of that colour divided by the total number (#12#) of socks. Once one sock of that colour has been removed, the probability of the second sock being the same colour is the number of socks remaining of that colour divided by the remaining total (#11#) number of socks in the drawer.

So the probability of choosing any pair with matching colour is:

#1/11+1/22+5/33 = 6/66+3/66+10/66 = 19/66#
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Answer 2

The probability that Jenny chooses two socks of the same color is 7/18, or approximately 0.389.

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Answer 3

The probability that Jenny chooses two socks of the same color can be calculated by considering the possible combinations of selecting socks of the same color and dividing it by the total number of possible combinations when selecting any two socks.

Total number of socks = 4 blue + 3 white + 5 black = 12 socks

  1. Selecting two blue socks: Number of ways to choose 2 blue socks = 4C2 (combinations of choosing 2 from 4)

  2. Selecting two white socks: Number of ways to choose 2 white socks = 3C2 (combinations of choosing 2 from 3)

  3. Selecting two black socks: Number of ways to choose 2 black socks = 5C2 (combinations of choosing 2 from 5)

Total number of combinations of selecting any two socks = 12C2 (combinations of choosing 2 from 12)

Now, the probability that Jenny chooses two socks of the same color is:

[ P(\text{same color}) = \frac{{\text{Number of ways to choose 2 socks of the same color}}}{{\text{Total number of combinations}}} ]

[ P(\text{same color}) = \frac{{4C2 + 3C2 + 5C2}}{{12C2}} ]

You can calculate the combinations using the formula (nCk = \frac{{n!}}{{k! \cdot (n-k)!}}), where (n!) represents the factorial of (n), (k!) represents the factorial of (k), and ((n-k)!) represents the factorial of (n-k).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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