Question

An urn contains 17 white balls and 20 red balls. A sample of four balls is selected at random from the urn. What is the probability that the sample contains two white balls and two red ones

Answer 1

The probability that the sample of four balls contains two white balls and two red balls is 0.39.

Let E be an event a sample of selecting four balls from an urn are two red balls and two white balls.

According to the given question.

Total number of marbles in an urn = 17 + 20 = 37

Total number of white marbles = 17

Total number of red marbles = 20

Now, the number of ways of selecting four marbles form an urn = $^{37} C_{4}$

And, number of ways of selecting 2 white and two red marbles = $^{17} C_{2} \times\ ^{20} C_{2}$

As, we know that probability of an event is calculated by taking the ratio of total number of favorable outcomes to the total number of outcomes.

Therefore, the probability that the sample of four balls contains two white balls and two red balls is given by

$P(E) = \frac{^{20}C_{2} \times ^{17}C_{2} }{^{37}C_{4} }$

$\implies P(E) = \frac{\frac{20!\times 17!}{2!18!(2!!15!)} }{\frac{37!}{4!33!} }$

$\implies P(E) = \frac{\frac{20(19)(17)(16)}{2(2)} }{\frac{37(36)(35)(34)}{(4)(3)(2)} }$

⇒ P(E) = 5(19)(17)(16)/(37(3)(35)(17))

⇒ P(E) = 19(16)/37(3)(7)

⇒ P(E) = 304/777

⇒ P(E) = 0.39

Hence, the probability that the sample of four balls contains two white balls and two red balls is 0.39.

Find out more information about probability here:

brainly.com/question/11234923

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