Answer:
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
[tex]\frac{32*31*30*29*28*27*26*25*24*23*22*21*20}{52*51*50*49*48*47*46*45*44*43*42*41*40} = 0.000547[/tex]
So the probability that a randomly selected bridge hand is a Yarborough is
1 - 0.000547 = 0.99945
