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-BARSIC- [3]
3 years ago
6

If a single card is drawn from a standard 52-card deck, in how many ways could it be a diamond or a face card (a face card is a

Jack, Queen, or King)?
Mathematics
2 answers:
nataly862011 [7]3 years ago
6 0
There are 13 diamonds, and 4*3=12 face cards, of which 3 are diamonds.
So total of "diamond or face card" is 13+12-3=22.
=> 
Probability of drawing a diamond OR a face card from a full 52-card deck is
22/52=11/26.
zhuklara [117]3 years ago
4 0

Answer:

\frac{11}{26}

Step-by-step explanation:

Total number of cards in the deck = 52

Total number of face cards = 12

Total number of diamond cards = 13

Total number of diamond face card = 3

So, total number of ways become = 12+13-3=22

These are out of 52.

Hence, it could be a diamond or a face card = \frac{22}{52}=\frac{11}{26}

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Eighteen telephones have just been received at an authorized service center. Six of these telephones are cellular, six are cordl
Karo-lina-s [1.5K]

Full Question

Eighteen telephones have just been received at an authorized service center. Six of these telephones are cellular, six are cordless, and the other six are corded phones. Suppose that these components are randomly allocated the numbers 1, 2, . . . , 18 to establish the order in which they will be serviced.

What is the probability that after servicing twelve of these phones, phones of only two of the three types remain to be serviced?

What is the probability that two phones of each type are among the first six serviced?

Answer:

a. 0.149

b. 0.182

Step-by-step explanation:

Given

Number of telephone= 18

Number of cellular= 6

Number of cordless = 6

Number of corded = 6

a.

There are 18C6 ways of choosing 6 phones

18C6 = 18564

From the Question, there are 3 types of telephone (cordless, Corded and cellular)

There are 3C2 ways of choosing 2 out of 3 types of television

3C2 = 3

There are 12C6 ways of choosing last 6 phones from just 2 types (2 types = 6 + 6 = 12)

12C6 = 924

There are 2 * 6C6 * 6C0 ways of choosing none from any of these two types of phones

2 * 6C6 * 6C0 = 2 * 1 * 1 = 2.

So, the probability that after servicing twelve of these phones, phones of only two of the three types remain to be serviced is

3 * (924 - 2) / 18564

= 3 * 922/18564

= 2766/18564

= 0.149

b)

There are 6C2 * 6C2 * 6C2 ways of choosing 2 cellular, 2 cordless, 2 corded phones

= (6C2)³

= 3375

So, the probability that two phones of each type are among the first six serviced is

= 3375/18564

= 0.182

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