Respuesta :
Answer:
The correct answer is:
A. 1/169 - The probability of drawing a king followed by a queen with a replacement.
B.1/442 - The probability of drawing a red ace followed by another red ace without replacement.
C.4/442 - The probability of drawing a 3 or 5 followed by 4 or 6, with replacement.
D.1/52 - The probability of drawing a spade followed by a jack of ant color with replacement.
Step-by-step explanation:
The other answer is incorrect. I did the test and got the answer wrong because of it.
Probability of an event is the measure of its chance of occurrence. The probabilities for the considered events are:
- P(Drawing a red ace, then another red ace without replacement) = 0.000754
- P(drawing a 3 or 5 followed by 4 or 6, with replacement) = 0.00148
- P(drawing a king followed by a queen with a replacement) = 0.00037
- P(drawing a spade followed by a jack of ant color with replacement)= 0.0089
How to calculate the probability of an event?
Suppose that there are finite elementary events in the sample space of the considered experiment, and all are equally likely.
[tex]P(E) = \dfrac{\text{Number of favorable cases}}{\text{Number of total cases}}[/tex]
Where favorable cases are those elementary events who belong to E, and total cases are the size of the sample space.
What is the rule of product in combinatorics?
If a work A can be done in p ways, and another work B can be done in q ways, then both A and B can be done in [tex]p \times q[/tex] ways.
Remember that this count doesn't differentiate between order of doing A first or B first then doing other work after the first work.
Thus, doing A then B is considered same as doing B then A
For the given case, mainly two things are occurring. One is drawing with replacement and one is drawing without replacement.
When the first draw is followed by replacement, the deck doesn't change. But when no replacement occurs after first draw, then the deck doesn't remain same as before. This makes changes in number of ways we can draw an item from it.
Number of ways 2 cards can be selected one by one with replacement is:
First card can be selected from 52 cards' deck in 52 ways.
Since that card is replaced, thus, number of cards doesn't change.
The second card, thus, can be drawn in 52 ways.
By the rule of product in combinatorics, we get:
Total number of drawing 2 cards one-by one from 52 cards's deck, with replacement is: [tex]52 \times 52 = 2704[/tex]
Number of ways 2 cards can be selected one-by one without replacement is:
First card in 52 ways.
Now since there are only 51 cards left as there was no replacement done, thus, second card can be drawn in 51 ways.
Thus,
Total number of drawing 2 cards one-by one from 52 cards's deck, without replacement is: [tex]52 \times 51 = 2652[/tex]
Getting the needed probabilities:
- 1. The probability of drawing a red ace followed by another red ace without replacement.
Drawing a red ace once can be done in only 2 ways as there are 2 red aces in a deck of cards. Since there is no replacement, the second red ace can be drawn in 1 way only. Thus,
Favorable number of cases = 2 times 1 = 2 (by product rule)
Total cases of drawing two cards one by one with replacement = 2652,
Thus, needed probability = 2/2652 = 0.000754 approx.
P(Drawing a red ace, then another red ace without replacement) = 0.000754
- 2. The probability of drawing a 3 or 5 followed by 4 or 6, with replacement.
There are 4 threes and 4 fives, thus total 8 likely cards for first case, thus, total 8 ways of drawing them.
Now replacement ensures that all cards in deck are back.
Now for second card, we need four or six, whose quantity is 8 too, thus,
by rule of product, number of favorable cases = 8 times 8 = 64
Total cases of drawing two cards one by one with replacement = 2704,
Thus, needed probability = 64/2704 = 0.00148 approx.
P(drawing a 3 or 5 followed by 4 or 6, with replacement) = 0.00148
- 3. The probability of drawing a king followed by a queen with a replacement.
There are four kings, so first card can be drawn in four ways which would give us favorable result.
Replacement makes deck same as before, thus, for second card, 4 queens are surely available. Thus, second card can be drawn in 4 favorable cases.
Total number of cases=4 times 4 = 16 (by product rule)
Total cases of drawing two cards one by one with replacement = 2704,
Thus, needed probability = 16/2704 = 0.00037 approx.
P(drawing a king followed by a queen with a replacement) = 0.00037
- 4. The probability of drawing a spade followed by a jack of ant color with replacement.
There are 12 spades. Thus, first card can be drawn in 12 ways.
There is replacement, which makes deck as it was before.
Ant color is red. For red colored cards, there are 2 jacks. Thus, second card can be drawn favorably in 2 ways.
Thus, by rule of product, number of favorable cases = 2 times 12 = 24
Total cases of drawing two cards one by one with replacement = 2704,
Thus, needed probability = 24/2704 = 0.0089 approx.
P(drawing a spade followed by a jack of ant color with replacement)= 0.0089
Thus,
The probabilities for the considered events are:
- P(Drawing a red ace, then another red ace without replacement) = 0.000754
- P(drawing a 3 or 5 followed by 4 or 6, with replacement) = 0.00148
- P(drawing a king followed by a queen with a replacement) = 0.00037
- P(drawing a spade followed by a jack of ant color with replacement)= 0.0089
Learn more about probabilities here:
https://brainly.com/question/1210781