Question #53170

A,B,C,D cut a pack of 52 cards successively in the order given. If a person who cuts a spade first receives rs.700, what are their respective expectations?
1

Expert's answer

2015-12-04T14:33:33-0500

Answer on Question #53170 – Math – Statistics and Probability

A,B,C,D cut a pack of 52 cards successively in the order given. If a person who cuts a spade first receives rs.700, what are their respective expectations?

Solution

The probability of the first spade for the first person is


P(A)=1352=0.25.P(A) = \frac{13}{52} = 0.25.


The respective expectation for the first person is


E(A)=xP(x)=7000.25=175.E(A) = x \cdot P(x) = 700 \cdot 0.25 = 175.


The probability of the first spade for the second person is


P(B)=P(BA)P(A)=1351(11352)=13(5213)5152=39514=13174.P(B) = P(B|\overline{A}) \cdot P(\overline{A}) = \frac{13}{51} \cdot \left(1 - \frac{13}{52}\right) = \frac{13 \cdot (52 - 13)}{51 \cdot 52} = \frac{39}{51 \cdot 4} = \frac{13}{17 \cdot 4}.


The respective expectation for the second person is


E(B)=xP(B)=70013174133.82.E(B) = x \cdot P(B) = 700 \cdot \frac{13}{17 \cdot 4} \approx 133.82.


The probability of the first spade for the third person is


P(C)=P(BA,B)P(A,B)=1350521352511351=135039523851=133850417=131917100.P(C) = P(B|\overline{A}, \overline{B}) \cdot P(\overline{A}, \overline{B}) = \frac{13}{50} \cdot \frac{52 - 13}{52} \cdot \frac{51 - 13}{51} = \frac{13}{50} \cdot \frac{39}{52} \cdot \frac{38}{51} = \frac{13 \cdot 38}{50 \cdot 4 \cdot 17} = \frac{13 \cdot 19}{17 \cdot 100}.


The respective expectation for the third person is


E(C)=xP(B)=700131917100101.71.E(C) = x \cdot P(B) = 700 \cdot \frac{13 \cdot 19}{17 \cdot 100} \approx 101.71.


The probability of the first spade for the fourth person is


P(D)=P(BA,B,C)P(A,B,C)=1349521352511351501350=1349395238513750=13491438173750==13491219173750.\begin{array}{l} P(D) = P(B|\overline{A}, \overline{B}, \overline{C}) \cdot P(\overline{A}, \overline{B}, \overline{C}) = \frac{13}{49} \cdot \frac{52 - 13}{52} \cdot \frac{51 - 13}{51} \cdot \frac{50 - 13}{50} = \frac{13}{49} \cdot \frac{39}{52} \cdot \frac{38}{51} \cdot \frac{37}{50} = \frac{13}{49} \cdot \frac{1}{4} \cdot \frac{38}{17} \cdot \frac{37}{50} = \\ = \frac{13}{49} \cdot \frac{1}{2} \cdot \frac{19}{17} \cdot \frac{37}{50}. \end{array}


The respective expectation for the fourth person is


E(D)=xP(D)=700131937491710076.80.E(D) = x \cdot P(D) = 700 \cdot \frac{13 \cdot 19 \cdot 37}{49 \cdot 17 \cdot 100} \approx 76.80.


Let F=F = "a person cuts a spade first", then F={700,p1=1/4700,p2=13/(174)700,p3=(1319)/(17100)700,p4=(131937)/(4917100)F = \begin{cases} 700, & p_1 = 1/4 \\ 700, & p_2 = 13/(17 \cdot 4) \\ 700, & p_3 = (13 \cdot 19)/(17 \cdot 100) \\ 700, & p_4 = (13 \cdot 19 \cdot 37)/(49 \cdot 17 \cdot 100) \end{cases}

The respective expectation is


E(F)=700p1+700p2+700p3+700p4=700(14+13174+131917100+1319374917100)=487.33.E(F) = 700p_1 + 700p_2 + 700p_3 + 700p_4 = 700 \cdot \left(\frac{1}{4} + \frac{13}{17 \cdot 4} + \frac{13 \cdot 19}{17 \cdot 100} + \frac{13 \cdot 19 \cdot 37}{49 \cdot 17 \cdot 100}\right) = 487.33.


Answer: 487.33.

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