Shush fhoncizbizxoabcobad
Answer:
Step-by-step explanation:
given that a deck of cards is shuffled.
we know in a deck there are 52 cards, 13 cards of each variety spade, clubs hearts and dice. Red are 26 and black are 26. kings, will be 4.
(a) the top card is the king of spades and the bottom card is the queen of spades?
(iii) 1/52 × 1/52
Top has 1/52 and bottom has 1/52 and these are independent.
(b) the top card is the king of spades and the bottom card is the king of spades?
(viii) None of the above
Because it is impossible.
(c) the top card is the king of spades or the bottom card is the king of spades?
(iv) 1/52 + 1/52
This is the sum of probabilities because there is no common event for these two.
(d) the top card is the king of spades or the bottom card is the queen of spades?
(ii) 1/52 + 1/51 (once king of spades is there, then probability is 1/51 for bottom card)
(e) of the top and bottom cards, one is the king of spades and the other is the queen of spades?
(vii) 2/52 × 1/51
Because this is twice of probability d.
Answer:
The probability of getting two of the same color is 61/121 or about 50.41%.
Step-by-step explanation:
The bag is filled with five blue marbles and six red marbles.
And we want to find the probability of getting two of the same color.
If we're getting two of the same color, this means that we are either getting Red - Red or Blue - Blue.
In other words, we can find the independent probability of each case and add the probabilities together*.
The probability of getting a red marble first is:

Since the marble is replaced, the probability of getting another red is: 
The probability of getting a blue marble first is:

And the probability of getting another blue is:

So, the probability of getting two of the same color is:

*Note:
We can only add the probabilities together because the event is mutually exclusive. That is, a red marble is a red marble and a blue marble is a blue marble: a marble cannot be both red and blue simultaneously.
Answer:
30
Step-by-step explanation:
Answer:
a) 
b) 
c) Mary's score was 241.25.
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:

a) Find the z-score of John who scored 190



b) Find the z-score of Bill who scored 270



c) If Mary had a score of 1.25, what was Mary’s score?




Mary's score was 241.25.