A lot of 4 white and 4 red balls is randomly divided into two halves.

1.	A lot of 4 white and 4 red balls is randomly divided into two halves. What is the probability that there will be 2 red and 2 white balls in each half?1. 18/352. 3/353. 1/24. None of the above
Answer» Correct Answer - Option 1 : 18/35 Concept: we know that \(\rm ^nC_r\) = \(\rm \dfrac{n!}{r!(n-r)!}\) Calculations: we are picking the four balls without replacement and that the 4 white and 4 red balls are the only balls in the box and that we choose 2 red balls and 2 white balls. Number of ways to choose 4 balls out of 8 = \(\rm ^8C_4\) = \(\rm \dfrac{8!}{4!4!}\) = 70 Number of ways to pick 2 red balls out of 4 = \(\rm ^4C_2\) = \(\rm \dfrac{4!}{2!2!}\) = 6 Number of ways to pick 2 white balls out of 4 = \(\rm ^4C_2\) = \(\rm \dfrac{4!}{2!2!}\) = 6 Probability = \(\frac {6 \times 6} {70} = \frac {36} {70} = \frac {18} {35} \)

A lot of 4 white and 4 red balls is randomly divided into two halves. What is the probability that there will be 2 red and 2 white balls in each half?1. 18/352. 3/353. 1/24. None of the above

Answer» Correct Answer - Option 1 : 18/35

Concept:

we know that

\(\rm ^nC_r\) = \(\rm \dfrac{n!}{r!(n-r)!}\)

Calculations:

we are picking the four balls without replacement and that the 4 white and 4 red balls are the only balls in the box and that we choose 2 red balls and 2 white balls.

Number of ways to choose 4 balls out of 8 = \(\rm ^8C_4\) = \(\rm \dfrac{8!}{4!4!}\) = 70

Number of ways to pick 2 red balls out of 4 = \(\rm ^4C_2\) = \(\rm \dfrac{4!}{2!2!}\) = 6

Number of ways to pick 2 white balls out of 4 = \(\rm ^4C_2\) = \(\rm \dfrac{4!}{2!2!}\) = 6

Probability = \(\frac {6 \times 6} {70} = \frac {36} {70} = \frac {18} {35} \)

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