Question

In the card game, bridge, played with an ordinary deck of 52 cards, all cards are dealt among four players, 13 each, randomly. What is the probability that each player gets one ace? Hint: Let A1 be the event that the ace of hearts is dealt to one of the four players. Let A2 be the event that the ace of hearts and ace of diamonds are dealt to two different players. Let A3 be the event that the ace of hearts, ace of diamonds, and ace of spades are dealt to three different players, and let A4 be the event that each player gets exactly one of the aces. Note that A4 ⊆ A3 ⊆ A2 ⊆ A1, and the desired probability is P(A4) = P(A1 ∩ A2 ∩ A3 ∩ A4). Use The Multiplication Rule in Conditional Probability.

          In the card game, bridge, played with an ordinary deck of 52 cards, all cards are dealt among four players, 13 each, randomly. What is the probability that each player gets one ace? Hint: Let A1 be the event that the ace of hearts is dealt to one of the four players. Let A2 be the event that the ace of hearts and ace of diamonds are dealt to two different players. Let A3 be the event that the ace of hearts, ace of diamonds, and ace of spades are dealt to three different players, and let A4 be the event that each player gets exactly one of the aces. Note that A4 ⊆ A3 ⊆ A2 ⊆ A1, and the desired probability is P(A4) = P(A1 ∩ A2 ∩ A3 ∩ A4). Use The Multiplication Rule in Conditional Probability.
        
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Elementary Statistics a Step by Step Approach
Elementary Statistics a Step by Step Approach
Allan G. Bluman 9th Edition
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In the card game, bridge, played with an ordinary deck of 52 cards, all cards are dealt among four players, 13 each, randomly. What is the probability that each player gets one ace? Hint: Let A1 be the event that the ace of hearts is dealt to one of the four players. Let A2 be the event that the ace of hearts and ace of diamonds are dealt to two different players. Let A3 be the event that the ace of hearts, ace of diamonds, and ace of spades are dealt to three different players, and let A4 be the event that each player gets exactly one of the aces. Note that A4 ⊆ A3 ⊆ A2 ⊆ A1, and the desired probability is P(A4) = P(A1 ∩ A2 ∩ A3 ∩ A4). Use The Multiplication Rule in Conditional Probability.
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Transcript

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00:01 To solve this question, first of all we will let here x1, x2 and so on till xb, b are random sample.
00:24 And sample sizes here n is a from the poison distribution.
00:37 So now our parameter here will be as lambda.
00:46 So as example our x -i will be for the poison of lambda parameter and then we can find our probability for the event x is equal to small x which will be as when x is equal to 0 1 2 and so on we will get our value e to the power of minus lambda lambda to the power of x divided by factorial x and otherwise it will be equal to 0.
01:28 So now here we will reject our null hypothesis which is lambda is equal to 0 .5 if the observed observed sum as submission of i is equal to 1 to p for x i is greater than not equal to 8.
02:00 So now we will find here our critical region that will be as x such that summation of i is equal to 1 to 8 of x i greater than or equal to 8 and according to the additive property as per additive property of poison distribution we have our value for submission of i is equal to 1.
02:51 1 to 8 as our sample size is 8 here x i submit p 8 lambda and now we will let here u is equal to submission of i is equal to 1 to 8 x i let's say thus and under our null hypothesis is ho when ho is true.
03:26 If ho is true then we can write here u is equal to submission of i is equal to 1 to 8 x i similar to p 8 in the bracket 0 .5 so now we can see u is for poison of you and now the critical reason here as for the question our critical reason is u is greater than or equal to 8 and therefore we can say to find lambda now to find lambda our alpha will be able to p type to error and therefore p we will reject null hypothesis when when h -o is true and therefore we can write here p u is greater than or equal to 8 slash lambda is equal to 0 .5 so therefore we get u similar to poison of you and therefore our value for alpha will be equal to 1 minus probability of getting u less than 8 so that that is 1 minus probability of getting u less than or equal to 7 and this event can be finded by probability of getting u is equal to 0 added by probability of getting u is equal to 1 further till probability of getting u is equal to 7 you you have to compute all these event here and the events will be equal to 1 minus e to the power of minus 4 4 to the power of 7 divided by factorial 0 added by e to the power of minus 4, 4 to the power of 1 divided by factorial 1 added by e to the power of minus 4 4 to the power of 2 divided by factorial 2 and so on till the 7 term which is e to the power of minus 4 4 to the power of 7 divided by factorial 7.
06:14 So this can be calculated further as 1 minus minus within the bracket we can take e to the power of minus 4 common here.
06:26 So it will be e to the power of minus 4 within the bracket when we saw this will be 1 plus 4 plus 16 by 2, 32 by 3 plus 32 by 3 plus 128 by 15 plus 256 divided by 455...
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