A box has 4 dice in it. 4 of them are fair and one of them is marked with 5 on all its faces. A dice chosen at random and rolled thrice shows 5 on all the occasions. The probability that the die chosen was a biased one is?

216/219

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215/220

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211/219

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214/220

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Solution

The correct option is A
216/219

Let X: Die shows up number 5

Y:It is an unfair die= $\frac{1}{4}$

Z:it is a normal die= $\frac{3}{4}$

$P (\frac{X}{Y}) = 1$

$P (\frac{X}{Z}) = \frac{1}{216}$

$P (\frac{Y}{X}) = \frac{\frac{1}{4} \times 1}{(\frac{1}{4} \times 1 + \frac{3}{4} \times \frac{1}{216})} = \frac{216}{219}$

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A box has 4 dice in it. 4 of them are fair and one of them is marked with 5 on all its faces. A dice chosen at random and rolled thrice shows 5 on all the occasions. The probability that the die chosen was a biased one is?

The correct option is A 216/219 Let X: Die shows up number 5 Y:It is an unfair die=14 Z:it is a normal die=34 P(XY)=1 P(XZ)=1216 P(YX)=14×1(14×1+34×1216)=216219