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Leya [2.2K]
1 year ago
8

How many different sequences of 3 playing cards (from a single 52-card deck) exist?

Mathematics
1 answer:
hichkok12 [17]1 year ago
4 0

There 1,32,600 possibilities to withdrawn 3 different sequence of playing cards from 52 cards.

<h3>What is Permutations?</h3>

A permutation of a set is, loosely speaking, an arrangement of its members into a sequence or linear order, or if the set is already ordered, a rearrangement of its elements.

Here, total number of cards = 52

         Number of Withdrawn cards = 3

Possibilities of different sequences of 3 playing cards = 52 X 51 X 50

                                                                                          = 132600

Thus, there 1,32,600 possibilities to withdrawn 3 different sequence of playing cards from 52 cards.

Learn more about Permutations from:

brainly.com/question/1216161

#SPJ1

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Probability of not getting question 2 = P(B') = 1 - P(B) = 1 - 0.80 = 0.20

To obtain the better option using the expected value method.

E(X) = Σ xᵢpᵢ

where pᵢ = each probability.

xᵢ = cash reward for each probability.

There are two ways to go about this.

Approach 1

If the contestant attempts question 1 first.

The possible probabilities include

1) The contestant misses the question 1 and cannot answer question 2 = P(A') = 0.40; cash reward associated = $0

2) The contestant gets the question 1 and misses question 2 = P(A n B') = P(A) × P(B') = 0.6 × 0.2 = 0.12; cash reward associated with this probability = $200

3) The contestant gets the question 1 and gets the question 2 too = P(A n B) = P(A) × P(B) = 0.6 × 0.8 = 0.48; cash reward associated with this probability = $300

Expected reward for this approach

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Approach 2

If the contestant attempts question 2 first.

The possible probabilities include

1) The contestant misses the question 2 and cannot answer question 1 = P(B') = 0.20; cash reward associated = $0

2) The contestant gets the question 2 and misses question 1 = P(A' n B) = P(A') × P(B) = 0.4 × 0.8 = 0.32; cash reward associated with this probability = $100

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Expected reward for this approach

E(X) = (0.2×0) + (0.32×100) + (0.48×300) = $176

Approach 2 is the better approach to follow as it has a higher expected reward.

The contestant should try and answer question 2 first to maximize the expected reward.

Hope this helps!!!

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