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USPshnik [31]
3 years ago
12

Two cards are dealt at random from a standard 52 card deck (without replacement). (Ace, King, Queen, Jack are face cards.)

Mathematics
1 answer:
Tema [17]3 years ago
7 0

Answer:

The answer is below

Step-by-step explanation:

There are 52 cards in a deck, 12 of these cards are face cards (4 kings, 4 queens and 4 jacks) and 40 are not face cards

a. Find the probability that the first card is a face card and the second is NOT a face card.

There are 12 first card, the probability that the first card is a face card is 12/52.

Since there are no replacement, after picking 1 face card the number of cards remaining is 51, the probability of the second card not being a face card = 40/51. Therefore:

The probability that the first card is a face card and the second is NOT a face card = P(first is face card) × P(second is not face card)  = 12/52 × 40/51 = 40/221

b) Find the probability that they are both face cards.

The probability that the first card is a face card is 12/52.

Since there are no replacement, after picking 1 face card the number of cards remaining is 51 and the number of face card remaining is 11, the probability of the second card is a face card = 11/51. Therefore:

The probability that they are both face cards = P(first is face card) × P(second is face card)  = 12/52 × 11/51 = 11/221

c) Find the probability that the second is a face card given the first is NOT a face card.

The probability that the first card is not a face card = 40/52

Since there are no replacement, after picking the first card the number of cards remaining is, the probability of the second card is a face card = 12/51. Therefore:

The probability that the second is a face card given the first is NOT a face card = P(first is not a face card) × P(second is face card)  = 40/52 × 12/51 = 40/221

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kodGreya [7K]

Given:-   r(t)=< at^2+1,t>  ; -\infty < t< \infty , where a is any positive real number.

Consider the helix parabolic equation :  

                                              r(t)=< at^2+1,t>

now, take the derivatives we get;

                                            r{}'(t)=

As, we know that two vectors are orthogonal if their dot product is zero.

Here,  r(t) and r{}'(t)  are orthogonal i.e,   r\cdot r{}'=0

Therefore, we have ,

                                  < at^2+1,t>\cdot < 2at,1>=0

< at^2+1,t>\cdot < 2at,1>=

                                              =2a^2t^3+2at+t

2a^2t^3+2at+t=0

take t common in above equation we get,

t\cdot \left (2a^2t^2+2a+1\right )=0

⇒t=0 or 2a^2t^2+2a+1=0

To find the solution for t;

take 2a^2t^2+2a+1=0

The numberD = b^2 -4ac determined from the coefficients of the equation ax^2 + bx + c = 0.

The determinant D=0-4(2a^2)(2a+1)=-8a^2\cdot(2a+1)

Since, for any positive value of a determinant is negative.

Therefore, there is no solution.

The only solution, we have t=0.

Hence, we have only one points on the parabola  r(t)=< at^2+1,t> i.e <1,0>




                                               




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