Question

. The adjacent figure shows a disc on which a player
spins an arrow twice. The fraction is formed, where ab is the number of sectors on which the arrow stops on the
first spin and is the number of sectors in which the
arrow stops on the second spin. On each spin, each sector
has equal chance of selection by the arrows. Find the 5
probability that fraction >1. (CBSE 2016)​

Answer 1

$\Large{\underline{\underline{\mathfrak{AnSwEr :}}}}$

$\sf{For \: \dfrac{a}{b} > 1, when \: = 1 \:, b \: cannot \: take \: any \: value.} \\ \\ \sf{ \ \ \ \ \ \ \ \ \ \ \ \ a = 2, b \: can \: take \: 1 \: value} \\ \\ \sf{ \ \ \ \ \ \ \ \ \ \ \ \ a = 3, b \: can \: take \: 2 \: value} \\ \\ \sf{ \ \ \ \ \ \ \ \ \ \ \ \ a = 4, b \: can \: take \: 3 \: value} \\ \\ \sf{ \ \ \ \ \ \ \ \ \ \ \ \ a = 5, b \: can \: take \: 4 \: value} \\ \\ \sf{ \ \ \ \ \ \ \ \ \ \ \ \ a = 6, b \: can \: take \: 5 \: value}$

$\rule{150}{2}$

$\sf{\dashrightarrow Total \: Possible \: outcomes = 36} \\ \\ \sf{\dashrightarrow P \bigg(\dfrac{a}{b} > 1 \bigg) = \dfrac{ 1+ 2 + 3+ 4 + 5}{36}} \\ \\ \sf{\dashrightarrow \bigg(\dfrac{a}{b} > 1 \bigg) = \dfrac{15}{36}} \\ \\ \sf{\dashrightarrow \bigg(\dfrac{a}{b} > 1 \bigg) = \dfrac{5}{12}}$

Answer 2

Answer:

no. of spins = 2 .

ab = 6 ( first spins)

ab = 6 (second spin)

☆ total outcomes in 2 spins = 6 + 6 = 12 .

● no. of outcomes ( >1) = 5 .

● 5 probability will be :-

$\frac{5}{12}$

5/12 is the same probability followed by 5 times .