A key to room number C 3 is dropped into a jar with five other keys, and the jar is throughly mixed. If keys are randomly drawn from the jar without replacement until the key to room C 3 is chosen, then what are the odds in favour that the key to room C 3 will be obtained on the 2 text nd try?

Question

A key to room number C 3 is dropped into a jar with five other keys, and the jar is throughly mixed. If keys are randomly drawn from the jar without replacement until the key to room C 3 is chosen, then what are the odds in favour that the key to room C 3 will be obtained on the 2 text nd try? (A) 1: 4 (B) 1: 5 (C) 1: 6 (D) 5: 6

Tardigrade Admin · Accepted Answer

We want to fail the first try, so we have (5/6) ⋅ (1/5)=(1/6) for the probability. The odds are therefore 1:5

$1 : 4$

$1 : 5$

$1 : 6$

$5 : 6$

We want to fail the first try, so we have $\frac{5}{6} \cdot \frac{1}{5} = \frac{1}{6}$ for the probability. The odds are therefore $1 : 5$

Q. A key to room number C3​ is dropped into a jar with five other keys, and the jar is throughly mixed. If keys are randomly drawn from the jar without replacement until the key to room C3​ is chosen, then what are the odds in favour that the key to room C3​ will be obtained on the 2nd try?

1:4

1:5

1:6

5:6

We want to fail the first try, so we have 65​⋅51​=61​ for the probability. The odds are therefore 1:5

$1 : 4$

$1 : 5$

$1 : 6$

$5 : 6$

We want to fail the first try, so we have $\frac{5}{6} \cdot \frac{1}{5} = \frac{1}{6}$ for the probability. The odds are therefore $1 : 5$