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Question

# A die is tossed twice. Find the probability of getting a number greater than 3 on each toss.

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Solution

## $S=\left\{\left(11\right),\left(12\right),\left(13\right),\left(14\right),\left(15\right),\left(16\right),\phantom{\rule{0ex}{0ex}}\left(21\right),\left(22\right),\left(23\right),\left(24\right),\left(25\right),\left(26\right),\phantom{\rule{0ex}{0ex}}\left(31\right),\left(32\right),\left(33\right),\left(34\right),\left(35\right),\left(36\right),\phantom{\rule{0ex}{0ex}}\left(41\right),\left(42\right),\left(43\right),\left(44\right),\left(45\right),\left(46\right),\phantom{\rule{0ex}{0ex}}\left(51\right),\left(52\right),\left(53\right),\left(54\right),\left(55\right),\left(56\right),\phantom{\rule{0ex}{0ex}}\left(61\right),\left(62\right),\left(63\right),\left(64\right),\left(65\right),\left(66\right)\right\}\phantom{\rule{0ex}{0ex}}n\left(\mathrm{S}\right)=36\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}E=\text{Getting a number greater than 3 on each toss}\phantom{\rule{0ex}{0ex}}=\left\{\left(44\right),\left(45\right),\left(46\right),\left(54\right),\left(55\right),\left(56\right),\left(64\right),\left(65\right),\left(66\right)\right\}\phantom{\rule{0ex}{0ex}}n\left(\mathrm{E}\right)=9\phantom{\rule{0ex}{0ex}}P\left(E\right)=\frac{9}{36}\phantom{\rule{0ex}{0ex}}=\frac{1}{4}$

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