What is the probability without replacement of drawing at least one ten (52 card deck)?
We need to know how many times we are drawing the cards ? we can answer it after knowing it.
thanks ,
Prasoon dixit
tutor maths
To find the probability of drawing at least one ten from a deck of 52 cards without replacement, we can break down the problem into two parts:
1. Finding the probability of not drawing any tens.
2. Subtracting the probability from 1 to find the probability of drawing at least one ten.
Let's start with the first part:
The deck contains 4 tens (10 of hearts, 10 of spades, 10 of diamonds, and 10 of clubs) out of 52 cards.
When we draw the first card, there are 52 options, and only 4 of them are tens. So the probability of not drawing a ten on the first draw is (52-4)/52 = 48/52 = 12/13.
Now, for the second draw, we have 51 cards left, and since we didn't draw any tens on the first draw, there are 3 tens left in the deck. So the probability of not drawing a ten on the second draw is (51-3)/51 = 48/51.
Continuing this logic, for the third draw, we have 50 cards left, and 2 tens. The probability of not drawing a ten is (50-2)/50 = 48/50 = 24/25.
Finally, for the fourth draw, we have 49 cards left, and only 1 ten. The probability of not drawing a ten is (49-1)/49 = 48/49.
To find the probability of not drawing any tens, we multiply the probabilities of each individual draw:
(12/13) * (48/51) * (24/25) * (48/49) = 0.6543
Now, we can find the probability of drawing at least one ten by subtracting this probability from 1:
1 - 0.6543 = 0.3457
So the probability of drawing at least one ten from a 52-card deck without replacement is approximately 0.3457, or about 34.57%.