SOLUTION: Dear tutor, Please help me solving these problems 1) An unbiased die is tossed. Find the probability that the number obtained is: (a) An event number (b) Less than 4 (c) Great

Question 1172247: Dear tutor, Please help me solving these problems
1) An unbiased die is tossed. Find the probability that the number obtained is:
(a) An event number (b) Less than 4 (c) Greater than 6
2) A card is randomly drawn from a standard deck of 52 playing cards. Find the probability that the card drawn is:
(a) A king (b) A diamond (c) Not a diamond (d) A king of diamonds
3) A letter is randomly chosen from the word NATIONAL. What is the probability that it will be?
(a) The letter A (b) A vowel (c) The letter S
4) A box contains 3 red discs, 4 green discs, and 5 blue discs. One disc is randomly drawn from the box. What is probability that will be:
(a) red disc (b) A green disc (c) A red disc or green disc (d) Not a blue disc

Found 2 solutions by math_tutor2020, ikleyn:
Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!
You should only post one question at a time. I'll do the first 2 problems to get you started.

===========================================

Problem 1

Part (a)

There are 3 even numbers (2,4,6) out of 6 total (1 through 6).
So 3/6 = 1/2 is the probability of rolling an even number.
Or you could say "half of the values are even".

Answer: 1/2

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Part (b)

We could use the same idea as part (a) above, but let's take a more formal approach. This approach is likely what many math textbooks use.

We'll use the concept of event space and sample space.
The event space is the set of all desired outcomes.
The sample space is the set of all possible outcomes (desired or not).
The event space is a subset of the sample space, which means that anything in set E is also in set S, but not vice versa.

Let's say
E = event space
S = sample space
both represent sets

Let n(E) represent the number of items in the event space.
Let n(S) represent the number of items in the sample space.

Dividing the numbers n(E) and n(S) leads to the probability of selecting an item randomly from the event space.
We can say
P(E) = n(E)/n(S)

With that bit of info in mind, we can tackle the problem.

E = {1,2,3} = set of values we want, everything less than 4
S = {1,2,3,4,5,6} = set of all possible values (some values in here are what we don't want)
everything in E is also in S, but not vice versa

n(E) = 3 = there are 3 items in set E
n(S) = 6 = there are 6 items in set S
P(E) = n(E)/n(S)
P(E) = 3/6
P(E) = 1/2
This approach may seem more tedious, but it's handy to get a formal structure of how probability is set up.

Answer: 1/2

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Part (c)

Event space = { } = empty set
Sample space = {1,2,3,4,5,6}
The event space is empty because there are no values greater than 6 on a standard die. This leads to a probability of 0 which says "the probability of getting a value greater than 6 is impossible".

More formally,
P(E) = n(E)/n(S)
P(E) = 0/6
P(E) = 0

Answer: 0

==============================================================================

Problem 2

Part (a)

Event space = {King of Hearts, King of Diamonds, King of Spades, King of Clubs}
Sample space = {52 cards in a standard deck}
We have 4 items in the event space, out of 52 items in the sample space.
We get 4/52 = 1/13.

Answer: 1/13

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Part (b)

There are 13 diamond cards out of 52 total.
We get the probability 13/52 = 1/4
You could also look at it like "there is one diamond suit out of 4 suits total".

Answer: 1/4

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Part (c)

We'll use the result from part (b). Subtract it from 1
This works because both events "getting diamonds" and "getting not diamonds" are complementary.
You either get a diamond or you don't.
P(not diamonds) = 1 - P(diamonds)
P(not diamonds) = 1 - 1/4
P(not diamonds) = 3/4
We can confirm this by noting there are 3 suits that aren't diamond out of 4 suits total.

Answer: 3/4

-----------

Part (d)

There's one card that is a king of diamonds out of 52 total.

Answer: 1/52
Answer by ikleyn(52781) (Show Source): You can put this solution on YOUR website!
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