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beks73 [17]
2 years ago
14

A game between two equally skilled players is ended before a winner is declared. Player A needs 3 more points to win, and player

B needs 2 more points to win. How should the stakes be divided between the two players? Blaise Pascal used the Arithmetical Triangle to solve this problem. In the division of the stakes, what is the ratio of player A's winnings to Player B's winnings?
Mathematics
1 answer:
alisha [4.7K]2 years ago
8 0

Answer: player A = 11/16 and player B = 5/16

Step-by-step explanation:

If a coin was to be tossed to determine the winner possible outcomes using arithmetical triangle.

Player A needs 2 points = 11

Player B needs 3 points = 5

Total outcome of tossing a coin = 16

Player A = 11/16 = 0.6875

Player B = 5/16 = 0. 3125

Or

Using the fifth roll of the pascal triangle (2+3) outcome

( 1, 4, 6, 4, 1 )

Addition of the first 3 represent Player A chances of winning = ( 1+ 4 + 6 ) = 11

And the last two = ( 1 + 4 ) = 5 represents the chances of team B winning

Total number of outcome = ( 1 + 4 + 6 + 4 + 1 ) = 16

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Orange M&M’s: The M&M’s web site says that 20% of milk chocolate M&M’s are orange. Let’s assume this is true and set
SOVA2 [1]

Answer:

The correct option is (A).

Step-by-step explanation:

Let <em>X</em> = number of orange  milk chocolate M&M’s.

The proportion of orange milk chocolate M&M’s is, <em>p</em> = 0.20.

The number of candies in a small bag of milk chocolate M&M’s is, <em>n</em> = 55.

The event of an milk chocolate M&M being orange is independent of the other candies.

The random variable <em>X</em> follows a Binomial distribution with parameter <em>n</em> = 55 and <em>p</em> = 0.20.

The expected value of a Binomial random variable is:

E(X)=np

Compute the expected number of orange  milk chocolate M&M’s in a bag of 55 candies as follows:

E(X)=np

         =55\times 0.20\\=11

It is provided that in a randomly selected bag of milk chocolate M&M's there were 14 orange ones, i.e. the proportion of orange milk chocolate M&M's in a random bag was 25.5%.

This proportion is not surprising.

This is because the average number of orange milk chocolate M&M’s in a bag of 55 candies is expected to be 11. So, if a bag has 14 orange milk chocolate M&M’s it is not unusual at all.

All unusual events have a very low probability, i.e. less than 0.05.

Compute the probability of P (X ≥ 14) as follows:

P(X\geq 14)=\sum\limits^{55}_{x=14}{{55\choose x}0.20^{x}(1-0.20)^{55-x}}

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This probability is quite larger than 0.05.

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Step-by-step explanation:

"If p, then q and if p, then r" cannot be used to draw a conclusion using the law of syllogism.

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