Answer:
The probability of drawing the compliment of a king or a queen from a standard deck of playing cards = 0.846
Step-by-step explanation:
<u><em>Step(i):-</em></u>
Let 'S' be the sample space associated with the drawing of a card
n (S) = 52C₁ = 52
Let E₁ be the event of the card drawn being a king
![n( E_{1} ) = 4 _{C_{1} } = 4](https://tex.z-dn.net/?f=n%28%20E_%7B1%7D%20%29%20%3D%204%20_%7BC_%7B1%7D%20%7D%20%20%3D%204)
Let E₂ be the event of the card drawn being a queen
![n( E_{2} ) = 4 _{C_{1} } = 4](https://tex.z-dn.net/?f=n%28%20E_%7B2%7D%20%29%20%3D%204%20_%7BC_%7B1%7D%20%7D%20%20%3D%204)
But E₁ and E₂ are mutually exclusive events
since E₁ U E₂ is the event of drawing a king or a queen
<u><em>step(ii):-</em></u>
The probability of drawing of a king or a queen from a standard deck of playing cards
P( E₁ U E₂ ) = P(E₁) +P(E₂)
![= \frac{4}{52} + \frac{4}{52}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B4%7D%7B52%7D%20%2B%20%5Cfrac%7B4%7D%7B52%7D)
P( E₁ U E₂ ) = ![\frac{8}{52}](https://tex.z-dn.net/?f=%5Cfrac%7B8%7D%7B52%7D)
<u><em>step(iii):-</em></u>
The probability of drawing the compliment of a king or a queen from a standard deck of playing cards
![P(E_{1}UE_{2}) ^{-} = 1- P(E_{1} U E_{2} )](https://tex.z-dn.net/?f=P%28E_%7B1%7DUE_%7B2%7D%29%20%20%5E%7B-%7D%20%3D%201-%20P%28E_%7B1%7D%20U%20E_%7B2%7D%20%29)
![P(E_{1}UE_{2}) ^{-} = 1- \frac{8}{52}](https://tex.z-dn.net/?f=P%28E_%7B1%7DUE_%7B2%7D%29%20%20%5E%7B-%7D%20%3D%201-%20%5Cfrac%7B8%7D%7B52%7D)
![P(E_{1}UE_{2}) ^{-} = \frac{52-8}{52} = \frac{44}{52} = 0.846](https://tex.z-dn.net/?f=P%28E_%7B1%7DUE_%7B2%7D%29%20%20%5E%7B-%7D%20%3D%20%5Cfrac%7B52-8%7D%7B52%7D%20%3D%20%5Cfrac%7B44%7D%7B52%7D%20%3D%200.846)
<u><em>Conclusion</em></u>:-
The probability of drawing the compliment of a king or a queen from a standard deck of playing cards = 0.846
Surface area = area of 2 triangular sides + area of the roof + area of the base.
= 2 * 1/2 * 6 * 4 + 2 * 5*7 + 6 * 7
= 24 + 70 + 42
= 136 units^2
The third integer is 86, hope this helps :)