There are 5 yellow, 4 green, and 2 blue marbles in a hat. You pick 2 marbles from the hat. Marbles are not returned after they have been drawn.
Find p( the first marble is blue and the second marble is blue)
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The rule for probability without replacement is the probability in the first draw times the probability for the second draw.
The probability for the first draw is the number of blue/the total number of marbles.
There are 2 blue balls and 11 balls total (5 yellow + 4 green + 2 blue) the probability of selecting a blue is 2/11
The probability for the second draw is the number of blue balls remaining/ the total number of balls remaining. Assuming that the first ball drawn was blue that would leave 1 blue ball and 10 balls total. So the probability that the second ball drawn would be blue is 1/10.
The probability of both being blue would be 2/11 x 1/10 = 2/110 = .0181818... or about 1.8%
In this question we need to find the probability that the first marble picked and the second marble picked are blue.
The probability of this happening is the product of the probability of the first marble being blue multiplied by the probability of the second marble being blue.
Lets consider the case where the first marble is being picked: Here the total number of marbles is 5+4+2=11. The number of blue marbles is 2. Therefore the probability that the marble being picked is blue is 2/11.
Now we move to the case where the second marble is being picked. As one blue marble was already picked in the first case, the total number of marbles is 5+4+1=10 and there is one blue marble. Therefore the probability that a blue marble is picked is 1/10.
Now taking the product of the two probabilities we get: (2/11)*(1/10)= 2/110=1/55.
Therefore the probability that the first marble and the second marble are blue is 1/55.
There are tototally 5+4+2 = 11 marble.
The number of green marbles = 2.
We require the probabilty of drawing one after the other 2 green marbles (without replacement) .
Getting a green marble in the first instance = 2/11.
To get both marbles to be green in the 1st and 2nd draws, it requires the 1st ball to be green, and having drawn the green marble first instance, we are left with only 1 green marble among a total of remaining 10 marbles. So the chance of green in the 2nd draw is (2-1)/(11-1) = 1/10
So the joint probablity first and 2nd balls drawn are green(without replacement) = (2/11)* (1/10) = 12/110 = 1/55.
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