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10 Jul 2024, 07:01
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P: 5
Q: 3
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
48% (03:03) correct
52%
(03:20)
wrong
based on 582
sessions
History
Date
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Not Attempted Yet
The king of Trishkinda has specially designed an ivory dice. Six exotic stones – agate, jasper, chalcedony, sapphire, topaz, garnet – are embedded on the six sides of the dice, such that one side has exactly one type of stone and no two sides have the same type of stone. When the dice is rolled, the probability of the agate stone appearing is \(\frac{1}{6}\). Let there be values P, Q, x, y such that when the dice is thrown P times, the probability of the agate stone appearing all P times is x and when the dice is thrown Q times, the probability of the agate stone appearing all Q times is y. Furthermore, \(x < y\) and \(\frac{1}{x} + \frac{1}{y} = 7,992\).
In the table, select a value for P and a value for Q that are jointly consistent with the given information. Make only two selections, one in each column.
In the table, select a value for P and a value for Q that are jointly consistent with the given information. Make only two selections, one in each column.
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| P | Q | |
| 1 | ||
| 2 | ||
| 3 | ||
| 4 | ||
| 5 | ||
| 6 |
ShowHide Answer
Official Answer
P: 5
Q: 3
10 Jul 2024, 07:14
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Bunuel
Experts' Global Explanation:
Probability of agate stone appearing when the dice is rolled 1 time = (1/6)
Probability of agate stone appearing when the dice is rolled 2 times = (1/6) x (1/6) = (1/6)2
Probability of agate stone appearing when the dice is rolled P times = (1/6)^P = x
Probability of agate stone appearing when the dice is rolled Q times = (1/6)^Q = y
1/x + 1/y = 7992
1/(1/6)^P + 1/(1/6)^Q = 7992
6^P + 6Q = 7992
6^P + 6^Q = 7992
Taking 6Q as the common term:
6^Q(6^(P – Q) + 1) = 7992 (Equation I)
6^Q(1 + 6^(P – Q)) = 7992
6^Q(1 + 6^(P – Q)) = 6 x 6 x 6 x 37
6^Q(1 + 6^(P – Q)) = 6 x 6 x 6 x (1 + 36)
6^Q(1 + 6^(P – Q)) = 6^3 x (1 + 6^2)
Comparing both sides:
Q = 3
P – Q = 2
P – 3 = 2
P = 5
Note that if we had taken 6^P as the common term in Equation I, we would have received P = 3 and Q =5 as the answer. Both (P = 5, Q = 3) and (P = 3 and Q =5) satisfy Equation I, however since the question specifies that x < y, it follows that (1/6)^P < (1/6)^Q, and thus P =5 and Q =3 is the only valid solution.
Hence, for “P” column, “5”, and for “Q” column, “3” is the correct combination of the answer choices.
10 Jul 2024, 07:38
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7992 = (6^3) x 37
37=36+1
36=6^2
So, we can write 7992 = (6^3) x (6^2 + 1) = 6^5 + 6^3 equation (i)
x=1/(6^P) and y=1/(6^Q)
We are given that x<y, so we can conclude that Q<P.
Using equation (i), we can assign the values.
P=5 and Q=3
37=36+1
36=6^2
So, we can write 7992 = (6^3) x (6^2 + 1) = 6^5 + 6^3 equation (i)
x=1/(6^P) and y=1/(6^Q)
We are given that x<y, so we can conclude that Q<P.
Using equation (i), we can assign the values.
P=5 and Q=3
