Question

An urn contains 11 white balls and 5 green balls. A sample of seven is selected at random. What is the probability that the sample contains at least one green ball

Answer 1

The probbaility that the sample conatins at least one green ball is 0.90.

Let E be the event that denotes that the sample conatins at least one green.

According to the given question.

Total number of balls in an urn = 11 + 5 = 16

Total number of white balls = 11

Total number so green balls = 5

Now, the number of ways of selecting 7 balls from 16 balls

= $^{16} C_{7}$

= 16!/(7!9!)

= 11440

And the total number of ways of selecting 7 balls which contain atleast on green ball = $^{11} C_{6} \times \ ^{5} C_{1} + ^{11} C_{5} \times \ ^{5} C_{2} + ^{11} C_{4} \times \ ^{5} C_{3} +^{11} C_{3} \times \ ^{4} C_{4}+^{11} C_{2} \times \ ^{5} C_{5}$

= $\frac{11!}{6!5!} \times \frac{5!}{1!4!} +\frac{5!}{2!3!} \times\frac{11!}{5!6!} +\frac{11!}{7!4!} \times\frac{5!}{3!2!} +\frac{11!}{8!3!} \times\frac{4!}{4!0!} +\frac{11!}{9!2!} \times\frac{5!}{0!5!}$

= 452 × 5 + 10 × 452 + 330 × 10 + 165×1 + 55 ×1

= 2260 +  4520 +3300 + 165 + 55

= 10300

Thereofore, the probability that the sample conatins at least one green ball is given by

P(E) = (total number of ways of selecting 7 balls which contain atleast on green ball) / (the number of ways of selecting 7 balls from 16 balls )

⇒ P(E) = 10300/11440

⇒ P(E) = 0.90

Hence, the probbaility that the sample conatins at least one green ball is 0.90.

Find out more information about probability here:

brainly.com/question/11234923

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