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

welll its a very hard question but ill

step-by-step explanation:

first you take the 6 dollars and add 10 to it then add the 50 to that then multiply by 8 with xy and your answer will be

hope this

40 i don’t know lololol