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Updated on: 29 Jan 2018, 20:45
Originally posted by clarkkent0610 on 02 Nov 2012, 14:43.
Last edited by Bunuel on 29 Jan 2018, 20:45, edited 2 times in total.
Last edited by Bunuel on 29 Jan 2018, 20:45, edited 2 times in total.
Renamed the topic and added OA.
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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For each player's turn in a certain board game, a card is drawn. 3/4 of the cards in the deck are marked with a circle, and the remaining cards are marked with a square. If five players draw a card and then return it to the deck, what is the probability that at least four of the cards drawn are marked with a square?
A. 1/4^3
B. 1/4^4
C. 5*(1/4^3)
D. 1/4^5
E. (3/2)*(1/4^4)
A. 1/4^3
B. 1/4^4
C. 5*(1/4^3)
D. 1/4^5
E. (3/2)*(1/4^4)
02 Nov 2012, 15:03
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clarkkent0610
The probability of circle is 3/4 and the probability of square is 1-3/4=1/4.
We want the probability that out of 5 cards drawn 4 OR 5 are squares.
The probability of 4 squares or the probability of SSSSC, is \(P(SSSSC)=\frac{5!}{4!}*(\frac{1}{4})^4*(\frac{3}{4})=\frac{15}{4^5}\). We are multiplying by \(\frac{5!}{4!}=5\) since SSSSC scenario can occur in 5 ways: SSSSC, SSSCS, SSCSS, SCSSS, and CSSSS (number of premutations of 5 letters SSSSC out of which 4 S's are identcal);
The probability of 5 squares or the probability of SSSSS, is simply \(P(SSSSS)=(\frac{1}{4})^5\).
Therefore the overall probability is \(\frac{15}{4^5}+\frac{1}{4^5}=\frac{1}{4^3}\).
Answer: A.
P.S. Please indicate OA for PS problems.
General Discussion
29 Jan 2018, 20:25
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Hi Bunuel,
I did it this way, please correct me. idk what did I miss out
- 4 squares = 5C4 * (1/4)^4 * (3/4)^1
- 5 squares = 5C5 * (1/4)^5 * (3/4)^0
Total prob = (21/4) * (1/4)^4
I did it this way, please correct me. idk what did I miss out
Quote:
- 4 squares = 5C4 * (1/4)^4 * (3/4)^1
- 5 squares = 5C5 * (1/4)^5 * (3/4)^0
Total prob = (21/4) * (1/4)^4
