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Byju's Answer

Standard XII

Mathematics

Axiomatic Approach

7 girls G1,...

Question

7 girls $G_{1}, G_{2}, G_{3}, . . . G_{7}$ are such that their ages are in order $G_{1} < G_{2} < G_{3} < . . . . . < G_{7}$ . Five girls are selected at random and arranged in increasing order of their ages, then the probability that $G_{5}$ & $G_{7}$ are not consecutive, is

\frac{17}{21}

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\frac{19}{21}

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\frac{20}{21}

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\frac{13}{21}

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Solution

The correct option is D $\frac{17}{21}$
Required probability $= 1 P (G_{5}$ $& G_{7}$ are consecutive)
$= 1 - \frac{^{4} C_{3}}{^{7} C_{5}} = \frac{17}{21}$

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7 girls G1,G2,G3,...G7 are such that their ages are in order G1<G2<G3<.....<G7. Five girls are selected at random and arranged in increasing order of their ages, then the probability that G5 & G7 are not consecutive, is

The correct option is D 1721Required probability =1P(G5 &G7 are consecutive)=1−4C37C5=1721

7 girls $G_{1}, G_{2}, G_{3}, . . . G_{7}$ are such that their ages are in order $G_{1} < G_{2} < G_{3} < . . . . . < G_{7}$ . Five girls are selected at random and arranged in increasing order of their ages, then the probability that $G_{5}$ & $G_{7}$ are not consecutive, is

The correct option is D $\frac{17}{21}$
Required probability $= 1 P (G_{5}$ $& G_{7}$ are consecutive)
$= 1 - \frac{^{4} C_{3}}{^{7} C_{5}} = \frac{17}{21}$