Given that a loaded die is thrown such that
![\[\mathbf{P}\left( \mathbf{1} \right)\text{ }=\text{ }\mathbf{P}\left( \mathbf{2} \right)\text{ }=\text{ }\mathbf{0}.\mathbf{2},\text{ }\mathbf{P}\left( \mathbf{3} \right)\text{ }=\text{ }\mathbf{P}\left( \mathbf{5} \right)\text{ }=\text{ }\mathbf{P}\left( \mathbf{6} \right)\text{ }=\text{ }\mathbf{0}.\mathbf{1}\text{ }\mathbf{and}\text{ }\mathbf{P}\left( \mathbf{4} \right)\text{ }=\text{ }\mathbf{0}.\mathbf{3}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-4174be6da398129f0164c8571610db50_l3.png "Rendered by QuickLaTeX.com")
and die is thrown two times. Also given that:
A = same number each time and
B = Total score is
![\[10\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-7e13914576c8f07fa7780b4b1aae684d_l3.png "Rendered by QuickLaTeX.com")
or more.
So, P(A) =
![\[\left[ P\left( 1,\text{ }1 \right)\text{ }+\text{ }P\left( 2,\text{ }2 \right)\text{ }+\text{ }P\left( 3,\text{ }3 \right)\text{ }+\text{ }P\left( 4,\text{ }4 \right)\text{ }+\text{ }P\left( 5,\text{ }5 \right)\text{ }+\text{ }P\left( 6,\text{ }6 \right) \right]\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-fa6dc8136fb7c187ad843ffe22ab4b3d_l3.png "Rendered by QuickLaTeX.com")
=
![\[P\left( 1 \right).P\left( 1 \right)\text{ }+\text{ }P\left( 2 \right).P\left( 2 \right)\text{ }+\text{ }P\left( 3 \right).P\left( 3 \right)\text{ }+\text{ }P\left( 4 \right).P\left( 4 \right)\text{ }+\text{ }P\left( 5 \right).P\left( 5 \right)\text{ }+\text{ }P\left( 6 \right).P\left( 6 \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-3c83795284c4f661083ee87c65f3d935_l3.png "Rendered by QuickLaTeX.com")
=
![\[0.2\text{ }x\text{ }0.2\text{ }+\text{ }0.2\text{ }x\text{ }0.2\text{ }+\text{ }0.1\text{ }x\text{ }0.1\text{ }+\text{ }0.3\text{ }x\text{ }0.3\text{ }+\text{ }0.1\text{ }x\text{ }0.1\text{ }+\text{ }0.1\text{ }x\text{ }0.1\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-35c6a626b2bc486ac136afc2177e87fe_l3.png "Rendered by QuickLaTeX.com")
=
![\[0.04\text{ }+\text{ }0.04\text{ }+\text{ }0.01\text{ }+\text{ }0.09\text{ }+\text{ }0.01\text{ }+\text{ }0.01\text{ }=\text{ }0.20\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-fb835a8a135410088abd62a4b3406531_l3.png "Rendered by QuickLaTeX.com")
Now, B =
![\[\left[ \left( 4,\text{ }6 \right),\text{ }\left( 6,\text{ }4 \right),\text{ }\left( 5,\text{ }5 \right),\text{ }\left( 5,\text{ }6 \right),\text{ }\left( 6,\text{ }5 \right),\text{ }\left( 6,\text{ }6 \right) \right]\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-425fa6a36eb9e73f380493236e258132_l3.png "Rendered by QuickLaTeX.com")
![\[P\left( B \right)\text{ }=\text{ }\left[ P\left( 4 \right).P\left( 6 \right)\text{ }+\text{ }P\left( 6 \right).P\left( 4 \right)\text{ }+\text{ }P\left( 5 \right).P\left( 5 \right)\text{ }+\text{ }P\left( 5 \right).P\left( 6 \right)\text{ }+\text{ }P\left( 6 \right).P\left( 5 \right)\text{ }+\text{ }P\left( 6 \right).P\left( 6 \right) \right]\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-9f59713c9adbb5303dfe35a9c62c671d_l3.png "Rendered by QuickLaTeX.com")
=
![\[0.3\text{ }x\text{ }0.1\text{ }+\text{ }0.1\text{ }x\text{ }0.3\text{ }+\text{ }0.1\text{ }x\text{ }0.1\text{ }+\text{ }0.1\text{ }x\text{ }0.1\text{ }+\text{ }0.1\text{ }x\text{ }0.1\text{ }+\text{ }0.1\text{ }x\text{ }0.1\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-2c5e84a8b2ce7700c91156c285331935_l3.png "Rendered by QuickLaTeX.com")
=
![\[0.03\text{ }+\text{ }0.03\text{ }+\text{ }0.01\text{ }+\text{ }0.01\text{ }+\text{ }0.01\text{ }+\text{ }0.01\text{ }=\text{ }0.10\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-efa39b3f5137c4cc2932f2767d1b0f88_l3.png "Rendered by QuickLaTeX.com")
A and B both events will be independent if
![\[P(A\cap ~B)\text{ }=\text{ }P\left( A \right).P\left( B \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-010d20b777170a03ae43cbb33073a3cf_l3.png "Rendered by QuickLaTeX.com")
…. (i)
And, here
![\[(A\cap ~B)\text{ }=\text{ }\left\{ \left( 5,\text{ }5 \right),\text{ }\left( 6,\text{ }6 \right) \right\}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-c7c3574131ef7726885e28d56f3d7e10_l3.png "Rendered by QuickLaTeX.com")
So,
![\[P(A\cap B)\text{ }=\text{ }P\left( 5,\text{ }5 \right)\text{ }+\text{ }P\left( 6,\text{ }6 \right)\text{ }=\text{ }P\left( 5 \right).P\left( 5 \right)\text{ }+\text{ }P\left( 6 \right).P\left( 6 \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-88a9c47afd1ae368e551d047b11ddf57_l3.png "Rendered by QuickLaTeX.com")
=
![\[0.1\text{ }x\text{ }0.1\text{ }+\text{ }0.1\text{ }x\text{ }0.1\text{ }=\text{ }0.02\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-c5439d73b94393f10e87545c2a6825c7_l3.png "Rendered by QuickLaTeX.com")
From equation (i) we get,
![\[0.02\text{ }=\text{ }0.20\text{ }x\text{ }0.10\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-ad7afd7f05c19c833abc20368a86b629_l3.png "Rendered by QuickLaTeX.com")
![\[0.02\text{ }=\text{ }0.02\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-b78316bcdfe47d9d45901cf1078481a0_l3.png "Rendered by QuickLaTeX.com")
Therefore, A and B are independent events.