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Sedaia [141]
3 years ago
10

An urn contains 3 red and 7 black balls. players a and b withdraw balls from the urn consecutively until a red ball is selected.

find the probability that a selects the red ball. (a draws the first ball, then b, and so on. there is no replacement of the balls drawn.)
Mathematics
1 answer:
Elza [17]3 years ago
7 0

An urn contains 3 red and 7 black balls (10 in total). The probability to withdraw by player a first ball red from the urn is 3/10. If he draws black ball (the probability then is 1-3/10=7/10) , then player b must withdraw second ball black. In the urn remain 9 balls (among them 6 black). The probability to choose black ball is 7/10·6/9. Then if player a select a red ball, the probability becomes 7/10·6/9·3/8 (only 8 balls left and 3 red among them) and the probability for the player a to seelct black ball is 7/10·6/9·5/8 (only 8 balls left and 5 black among them) and so on;

Player a: 3/10, 7/10·6/9·3/8, 7/10·6/9·5/8·4/7·3/6, 7/10·6/9·5/8·4/7·3/6·2/5·3/4.

Use the sum rule to calculate the total probability;

3/10+ 7/10·6/9·3/8+ 7/10·6/9·5/8·4/7·3/6+ 7/10·6/9·5/8·4/7·3/6·2/5·3/4

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agasfer [191]

Answer:

(10k - m)^{5}=100000k-50000k^{4}m+10000k^{3}m^{2}-1000k^{2}m^{3}+50km^{4}-m^{5}

Step-by-step explanation:

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- The rule of expand the binomial is:

(a+b)^{n}=(a)^{n}+nC1(a)^{n-1}(b)+nC2(a)^{n-2}(b)^{2}+nC3(a)^{n-3}(b)^{3}+...............+(b)^{5}

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∵ 5C3 = 10

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∵ 5C5 = 1

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