The second Earl of Yarborough is reported to have bet at odds of 1000 to 1 that a bridge hand of 13 cards would contain at least one card that is ten or higher. (By ten or higher we mean that a card is either a ten, a jack, a queen, a king, or an ace.) Nowadays, we call a hand that has no cards higher than 9 a Yarborough. What is the probability that a randomly selected bridge hand is a Yarborough?

0.99945

Step-by-step explanation:

The probability that a bridge hand of 13 cards would contain at least one card that is ten or higher would be the inverse of the probability that all of the 13 cards would not contain any 10 or higher, in other words all of them are 9 or lower. From 2 to 9 we have 8 cards of 4 suits like that, so a total of 32 out of 52 cards

The probability of this to happen is

In the first draw, chance are 32/52

In the 2nd draw, chance are 31/51

In the 3rd draw, chance are 30/50

...

In the 13th draw, chance are 20/40

So the total probability is

So the probability that a randomly selected bridge hand is a Yarborough is

1 - 0.000547 = 0.99945

a) one student is chosen from classroom a, then a student is chosen from classroom b.  

step-by-step explanation:

independent events are ones in which the probability of one event does not affect the probability of the second event.

one student is chosen from classroom a; this does not affect the chances of choosing a student from classroom b.   this makes these independent events.

when one card is chosen from a standard deck and then set aside before a second card is chosen, the probability of the second card being chosen changes.   this makes them dependent events.

when one student is chosen from classroom a and then a second student is chosen from classroom a, the probability of choosing the second student changes.   this makes them dependent events.

when a letter is chosen from the alphabet and then a second letter is chosen, the probability of choosing the second letter changes.   this makes them dependent events.

just take the area of the square and then the area of the circle then subtract the circle area from the squares area

step-by-step explanation: