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# A buys 3 tickets of a lottery in which total number of tickets is 12 out of which 3 carry prizes on them. A's chance of getting a prize is ;

$\begin {array} {1 1} (1)\;\large\frac{43}{55} & \quad (2)\;\large\frac{37}{55} \\ (3)\;\large\frac{34}{55} & \quad (4)\;None\: of \: these \end {array}$

We have P(A does not get prize ) = P ( all of A's tickets are blank )
$= \large\frac{9}{12} \times \large\frac{8}{11} \times \large\frac{7}{10} = \large\frac{21}{55}$
Hence P(A gets at least one prize ) = $1 - \large\frac{21}{55}$
$= \large\frac{55-21}{55}$
$= \large\frac{34}{55}$