Probability Help

preziano

New Member
Hi,

I giving my IGCSE this November and came across a problem that I can't solve. It'd be really helpful if someone can help me with the sum.

" A square board measuring 30 cm by 30 cm has two rectangular cards attached to it. The large card measures 15 cm by 20 cm and the small card measures 8 cm by 12 cm. The large card covers exactly one-quarter of the small card. If a dart is thrown at the board, what is the probability of piercing:

a) both cards
b) both cards, given that it pierces at least one of them"

I got the answer to question a) by finding the intersection

1/4 x [(8x12) / (30x30)] = 24 / 900 = 2/75

but cannot find anyway to solve question b).

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Con-Tester

Member
You’ll need to use the conditional probability formula. In the formula, outcome A = “dart pierces both cards”, and outcome B = “dart pierces at least one card”. The probabilities are directly related to the surface areas of the four possible regions (no card, card 1, card2, both cards) the dart can stick in.

(It is worth noting that the whole problem is already one of conditional probability because it implicitly assumes that the dart sticks somewhere in the 30 × 30 cm board.)

preziano

New Member
Thanks for the response, I cant seem to understand what the question truly means. What is the probability that the darts hits both the cards given that it pierces at least one of them???

Con-Tester

Member
Refer to the Wikipedia article I linked to (click on the blue “conditional probability” text). Do you understand conditional probability, specifically what the formula “P(A|B) = P(A ∩ B)/P(B)” means?

As I suggested earlier, use outcome A = “dart pierces both cards”, and outcome B = “dart pierces at least one card”. From your diagram, the following should be clear:

(1) P(A ∩ B) is the probability that the dart pierces both cards and it pierces at least one card. But if it pierces both, then it is also true that it pierces at least one of them. Therefore, outcome A subsumes outcome B, so that P(A ∩ B) = P(A), which you have already calculated as 24/900.

(2) P(B) is true if the dart hits any card or both of them. The total area covered by the cards is 20×15+(3/4)×12×8 cm² = 372 cm² out of 900 cm² of the board. Therefore, P(B) = 372/900.

Putting these values in the formula gives P(A|B) = P(A ∩ B)/P(B) = (24/900)/(372/900) = 24/372 = 2/31.

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Edit: What the question means is the following: Take all of the dart throws at the board. From them, select only those throws where the dart pierced at least one of the cards. In that subset, what proportion pierced both cards?

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dj_johnphillips

New Member
Thanks for the response, I cant seem to understand what the question truly means. What is the probability that the darts hits both the cards given that it pierces at least one of them???
Maybe this example will help?

My Aunt is going to come and visit me at some point next week, with an equal likelihood of any day.

Q. What is the probability that she will visit on a Wednesday?
A. 1/7

Q. What is the probability that she will visit on a Saturday?
A. 1/7

New information: My aunt has now told me she will be visiting on a weekday.

Q. What is the probability that she will visit on a Wednesday, given she will be visiting on a weekday?
A. 1/5

Q. What is the probability that she will visit on a Saturday, given she will be visiting on a weekday?
A. 0

The "given" in some way limits or clarifies the situation which the question is asking about.

preziano

New Member
That was really helpful. I kept dividing (2/75) [the answer of both cards] with (372/900) [any card or both]. Thanks a lot for clearing my doubt.

BGM

TS Contributor
Yes. As mentioned above, the information is "shrinking" the sample space from the 30 x 30 board to the union of two cards.