Let us combine 4 standard decks of cards (refer to question 3 for the definition

Question

Let us combine 4 standard decks of cards (refer to question 3 for the definition of the standard deck). That is, we will have 208 cards in total. The diamonds and hearts arc called red cards. If we randomly select 100 cards (without replacement), find the probability of getting exactly 47 red cards. If we randomly select a card, note its colour, and then put it back in the deck, and do this 100 times, find the probability of getting cither 43 or 44 red cards. Use a normal approximation to estimate the answer to (b).

Explanation / Answer

Given:

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Part a)

If we randomly select 100 cards, to find probability of 47 cards to be red

The probability distribution followed by such a sample is hypergeometric distribution

It is expressed as follows:

h(x;N,n,k) = [kCx] * [N-kCn-x] / [NCn]

Where : x = the exact number for which the probability is calculated

x = 47

N = total number so cards = 208

n = number so cards selected = 100

k = total number of red cards = 104

Hence,

104C47 * 104C53 / 208C100

h[47; 208, 100,104] = 0.07825................[Answer]

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Part b)

In this case since we replace back the card, the total number of cards for each trial remains same and hence the distribution followed becomes a Binomial one.

For a distribution to be a Binomial distribution following conditions must be met:

Hence the Binomial distribution probability for the given question can be given by the following formula:

b(x; n , P) = nCx * P^x * (1-p)^n-x

Where n = 100

p = 0.5

and x = 43 or 44

P(X=43 or 44) = P(x=43) + P(x=44)

P(X=43) = 100C43 * (0.5) ^43 * (0.5) ^ (100-43)

P(X=43) = 0.03

P(X= 44) = 100C44 * (0.5) ^44 * (0.5) ^ (100-44)

P(X= 44) = 0.039

Hence P(X= 43 or 44) = 0.03 + 0.039

P(X= 43 or 44) = 0.069....................[Answer]

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Part c)

Normal approximation estimate for part b can be applied as follows:

Step I : for normal approximation to be applied the following conditions must be met:

np > 5 and n(1-p) > 5

n = 100 , p = 0.5 , (1-0.5) = 0.5

Hence 100 *.5 = 50 > 5 and 100 * (1-0.5) = 50 > 5

Both the conditions are satisfied

Step II: Find mean and standard deviation:

Mean (µ) = np = 100 * 0.5 = 50

Standard deviation () = n*p*(1-p)

Standard deviation () = 100 * 0.5 * 0.5 = 5

Step III : apply continuity correction ( subtract and add 0.5 to the vlaue of x)

[This is done because we are converting a discrete distribution(Binomial distribution) to a continuous distribution (Normal distribution)]

P(42.5 <x < 44.5) = P(x< 44.5) - P(x<42.5)

Step IV: In order to find the probability we need to calculate the Z scores of each of the cases

Z1 for P(x<44.5) is

Z1 = (44.5 - 50)/ 5 = -1.1

Refer to the Z table to get the probability of Z = -1.1

P(Z<-1.1) = 0.1357

Z2 for P(x < 42.5) is

Z2 = (42.5-50)/5 = -1.5

Refer to the Z table to get the probability of Z = -1.5

P(Z<-1.5) = 0.0668

P(42.5<X<44.5) = 0.1357 - 0.0668

P(42.5<X<44.5) = 0.0689......................[Answer]

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