Cards are dealt, one at a time, from a standard 52 -card...

Name: Problem 75, Chapter 2: Mathematical Statistics with Applications
Uploaded: 2021-04-11T09:24:43-08:00
Duration: 3 min 50 s
Channel: Manisha Sarker
Description: Cards are dealt, one at a time, from a standard 52 -card deck. a. If the first 2 cards are both spades, what is the probability that the next 3 cards are also spades? b. If the first 3 cards are all spades, what is the probability that the next 2 cards are also spades? c. If the first 4 cards are all spades, what is the probability that the next card is also a spade?

Cards are dealt, one at a time, from a standard 52 -card deck. a. If the first 2 cards are both spades, what is the probability that the next 3 cards are also spades? b. If the first 3 cards are all spades, what is the probability that the next 2 cards are also spades? c. If the first 4 cards are all spades, what is the probability that the next card is also a spade?

Cards are dealt, one at a time, from a standard 52 -card deck.
a. If the first 2 cards are both spades, what is the probability that the next 3 cards are also spades?
b. If the first 3 cards are all spades, what is the probability that the next 2 cards are also spades?
c. If the first 4 cards are all spades, what is the probability that the next card is also a spade?

Key Concepts

Multiplication Rule for Dependent Events

The multiplication rule for dependent events states that the probability of a sequence of events occurring is the product of the probability of the first event and the conditional probabilities of each subsequent event given the preceding events. This rule is vital in sequential probability problems where earlier outcomes influence later probabilities.

Conditional Probability

Conditional probability refers to the probability of an event occurring given that another event has already occurred. This concept is essential when analyzing outcomes that depend on previous events, such as card draws where earlier outcomes affect the chances of later ones.

Sampling Without Replacement

Sampling without replacement is the process of drawing items from a finite set where each selected item is not returned to the set. This affects the probabilities of subsequent outcomes, as the composition of the set changes after each draw, making the events dependent.

Hypergeometric Distribution

The hypergeometric distribution models the probability of obtaining a certain number of successes in draws from a finite population without replacement. In the context of card games, it helps calculate the likelihood of drawing a specific number of desired cards (like spades) from the deck after some cards have already been drawn.

Transcript

00:01 In this problem, we are going to determine the probabilities of certain events.

00:05 Now in this problem, cards are dealt one at a time from a standard 52 card deck.

00:11 Now in the first problem, we need to determine the probability that if the first two cards are both spaced, then what is the probability that the next three cards are also spaced? so in order to find the probability, first of all, we need to find the number of favorable outcomes.

00:27 Now, in a standard 52 card deck there are 13 space.

00:30 And since the first two cards are both spades, that means out of these 13 spades, the number of spades left is 13 minus 2, which is 11.

00:39 And since we need to find the probability that the next three cards are spades, that means we need to choose three spades out of the remaining 11 space.

00:47 So the number of ways that can be done is equals to 11.

00:52 Now the total number of outcomes also needs to be determined.

00:55 There are 52 cards and two cards have already been drawn.

00:59 So that means that there are 50 cards remaining and out of that three cards need to be selected.

01:04 So that means the total number of outcomes will be 50 c3.

01:09 So if this is calculated, then 11c3 is equals to 165 and 50c3 is equals to 1960.

01:22 So 165 divided by 1960 0 is approximately equals to 0.

01:29 0 .0084.

01:32 So that is the required probability in this case.

01:36 Now in the second problem, if the first three cards are spaced, then what is the probability that the next two cards are also spades? so we first of all need to find the number of favorable outcomes.

01:47 Out of 13 spades, three spades have already been selected, so the remaining number of spades is 13 minus 3 equals to 10.

01:54 And out of that we need to find that the next two cards are spades.

01:58 So out of those 10 remaining spades, we need to select two spades...

Please give Ace some feedback