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R Ssays

Hi I have a question on number 40 Form C02 (June 2020). I thought that for a probability involving “or”, you do P(A) + P(B) – P(A and B). So in this case wouldn’t it be P(blue object) + P(triangle) – P(blue object and triangle)? So (3 blue objects/ 6 objects) + (3 triangles/ 6 objects) – (1/2*1/2) = 3/4? Or am I supposed to make P(blue object and triangle) = 1/6, making the final answer be 3/6 + 3/6 – 1/6 = 5/6? I was just confused on the P(blue object and triangle) since I originally thought you multiplied P(blue object)* P(triangle) to get that.

When they asked for the probability that it is a blue object or a triangle, then in this case I just counted the total number of objects that satisfy either of those two conditions, which is 5 here and so you can directly write the elementary probability of the event as 5/6.

You can apply the inclusion-exclusion principle, P(A or B) = P(A) + P(B) – P(A and B), and it is the same thing that I did. In this case, you are right that P(blue) = 3/6, P(triangle) = 3/6, and P(blue and triangle) = 1/6 (because there is only one object that is blue and a triangle), and we get the same final answer of 3/6 + 3/6 – 1/6=5/6.

Here the two events, picking a blue object and picking a triangle, are not independent and we can’t use the formula P(A and B) = P(A) times P(B).

That came from the equation of the circle that is given in the problem: $(x-2)^2 + y^2 = 1$, which gives us both the coordinates of the center and the radius. In general, if the equation of a circle is $(x-h)^2 + (y-k)^2=r^2$, then the center of the circle is at the point $(h, k)$ and the radius of the circle is $r$.

The 2020 June ACT test is Form C02. The form code C03 corresponds to 2019 December ACT. Do I have a mistake anywhere on this page that I am somehow overlooking?

R S says

Hi I have a question on number 40 Form C02 (June 2020). I thought that for a probability involving “or”, you do P(A) + P(B) – P(A and B). So in this case wouldn’t it be P(blue object) + P(triangle) – P(blue object and triangle)? So (3 blue objects/ 6 objects) + (3 triangles/ 6 objects) – (1/2*1/2) = 3/4? Or am I supposed to make P(blue object and triangle) = 1/6, making the final answer be 3/6 + 3/6 – 1/6 = 5/6? I was just confused on the P(blue object and triangle) since I originally thought you multiplied P(blue object)* P(triangle) to get that.

Dabral says

When they asked for the probability that it is a blue object or a triangle, then in this case I just counted the total number of objects that satisfy either of those two conditions, which is 5 here and so you can directly write the elementary probability of the event as 5/6.

You can apply the inclusion-exclusion principle, P(A or B) = P(A) + P(B) – P(A and B), and it is the same thing that I did. In this case, you are right that P(blue) = 3/6, P(triangle) = 3/6, and P(blue and triangle) = 1/6 (because there is only one object that is blue and a triangle), and we get the same final answer of 3/6 + 3/6 – 1/6=5/6.

Here the two events, picking a blue object and picking a triangle, are not independent and we can’t use the formula P(A and B) = P(A) times P(B).

Luke Johnson says

I have a question about number 24. To start the problem, you state that point B is at (2,0). I was wondering how you got to that conclusion?

Dabral says

Hi Luke,

That came from the equation of the circle that is given in the problem: $(x-2)^2 + y^2 = 1$, which gives us both the coordinates of the center and the radius. In general, if the equation of a circle is $(x-h)^2 + (y-k)^2=r^2$, then the center of the circle is at the point $(h, k)$ and the radius of the circle is $r$.

Dabral

Emily Bell says

Do you mean c02 or c03?

Dabral says

The 2020 June ACT test is Form C02. The form code C03 corresponds to 2019 December ACT. Do I have a mistake anywhere on this page that I am somehow overlooking?

Thanks

Eman saad says

Great work