Question #58971

An urn contains 4 white balls and 7 red balls four balls are selected. In how many ways can the 4 balls be drawn from the total of 11 ​balls:
​(a) If 3 balls are white and 1 is​ red?
​(b) If all 4 balls are​ white?
​(c) If all 4 balls are​ red?
1

Expert's answer

2016-04-08T09:46:05-0400

Answer on Question #58971 – Math – Statistics and Probability

Question

An urn contains 4 white balls and 7 red balls, four balls are selected. In how many ways can the 4 balls be drawn from the total of 11 balls:

(a) If 3 balls are white and 1 is red?

Solution

To find how many ways the 3 white balls can be drawn from the total 4 white balls we need to find the number of combinations without repetition:


C43=4!3!(43)!=43213211=4.C _ {4} ^ {3} = \frac {4 !}{3 ! (4 - 3) !} = \frac {4 \cdot 3 \cdot 2 \cdot 1}{3 \cdot 2 \cdot 1 \cdot 1} = 4.


To find how many ways the 1 red ball can be drawn from the total 7 red balls we need to find the number of combinations without repetition:


C71=7!1!(71)!=7!6!=7.C _ {7} ^ {1} = \frac {7 !}{1 ! (7 - 1) !} = \frac {7 !}{6 !} = 7.


So the total number of combinations:


N=C43C71=47=28.N = C _ {4} ^ {3} \cdot C _ {7} ^ {1} = 4 \cdot 7 = 28.


Answer: 28.

Question

An urn contains 4 white balls and 7 red balls four balls are selected. In how many ways can the 4 balls be drawn from the total of 11 balls:

(b) If all 4 balls are white?

Solution:

To find how many ways the 4 white balls can be drawn from the total 4 white balls we need to find the number of combinations without repetition:


C44=4!4!=1.C _ {4} ^ {4} = \frac {4 !}{4 !} = 1.


Answer: 1.

Question

An urn contains 4 white balls and 7 red balls four balls are selected. In how many ways can the 4 balls be drawn from the total of 11 balls:

(c) If all 4 balls are red?

Solution

To find how many ways the 4 red balls can be drawn from the total 7 red balls we need to find the number of combinations without repetition:


C74=7!4!(74)!=7!4!3!=765432432432=75=35.C _ {7} ^ {4} = \frac {7 !}{4 ! (7 - 4) !} = \frac {7 !}{4 ! \cdot 3 !} = \frac {7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2}{4 \cdot 3 \cdot 2 \cdot 4 \cdot 3 \cdot 2} = 7 \cdot 5 = 35.


Answer: 35.

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