I need help on some statistics problems. Could someone help me please?

Question

8. Solve the problem.
Four students drive to school in the same car. The students claim they were late to school and missed a test because of a flat tire. On the makeup test, the instructor asks the students to Identify the tire that went flat; front driver’s side, front passenger’s side, rear driver’s side, or rear passenger’s side. If the students didn’t really have a flat tire and each randomly selects a tire, what is the probability that all four students select the same tire?

A.1/8
B.1/4
C.1/64
D.1/256

9. Solve the problem.
Decide if the events A and B are mutually exclusive or not mutually exclusive. A card is drawn from a standard deck
of 52 playing cards.

*The result is a 7.
*The result is a jack.

A. mutually exclusive
B. not mutually exclusive

10. Solve the problem.
A card is drawn from a standard deck of 52 playing cards. Find the probability that the card is an ace or a black card.

A. 15/26
B. 29/52
C. 4/13
D. 7/13

Answer 1

8.) Suppose the students are interrogated one at a time. The first student can name any tire he wants. He doesn't have to match anything. The second student has a probability of 1/4th of naming the same tire as the first student. Similarly for the third and the fourth students. So the probability of them all naming the same tire is (1/4)(1/4)(1/4) = 1/64

9.) The set of 7's and the set of jacks are mutually exclusive, so the two events are mutually exclusive. You can't draw one card from the deck and have it be a 7 and a jack at the same time.

10.) There are 26 black cards and there are two aces that are not also black. That's 28 successful outcomes out of a total of 52. The probability of success is: 28/52 = 7/13.

Answer 2

#8
This is a binomial probability problem for which we use the following formula:
P(r) = nCr *p^r q^(n-r) In this case n =4, r=4, p=.25, q =.7
So, plugging the numbers in we have this:
4C4 *.25^4*.75^(4-4)
1 *.00390625 *1
=0.00390625
=1/256

#9 Yes, they are mutually exclusive. A card cannot be a 7 and a jack at the same time.

#10
P(A or B) = P(A) + P(B) –P(A∩B) Where ∩ means intersection.
=4/52 + 26/52 – 2/52
=4/52 +26/52 -2/52
=(4+26-2)/52
=28/52
=7/13

Hope this helps.
FE

Answer 3

Q8)

p(1st pupils choice) = 1/4 ( 4 tyres, pupil 1 picks one at random)

p(2nd pupils choice same as first) = 1/4

p(3rd pupils choice same as first) = 1/4

p(4th pupils choice same as first) = 1/4

p[(first pupil picking a tyre) AND (2nd pupil picking same tyre) AND (3rd pupil picking the same) AND (4th pupil picking the same tyre)

Using the AND rule which says when events are independent to find the combined probability, multiply the individual probabilities.

So p(all 4 pick the same) = 1/4 x 1/4 x1/4 x1/4 = 1/256

Q9

Mutually exclusive means events cannot happen together. It is impossible to pick a single card which is both a jack and a seven, so these events are mutually exclusive.


Q10,

There are 4 aces and 24 other cards that are black. (Two full suits are black hence 26 cards, but both aces have already been counted)

So the p( getting an Ace or a black card ) = 28/52 = 7/13