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Gun 1 and Gun 2 are shooting at the same target. Gun 1 shoots on the average nine shots during the same time Gun 2 shoots 10 shots. The precision of these two guns is not the same. - Business Mathematics and Statistics

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प्रश्न

Gun 1 and Gun 2 are shooting at the same target. Gun 1 shoots on the average nine shots during the same time Gun 2 shoots 10 shots. The precision of these two guns is not the same. On the average, out of 10 shots from Gun 2 seven hit the target. In the course of shooting the target has been hit by a bullet, but it is not known which Gunshot this bullet. Find the chance that the target was hit by Gun 2?

बेरीज

उत्तर

Let the events be defined as follows:

B = Event of hitting the target

E1 = Event of Gun 1 hit the target

E2 = Event of Gun 2 hit the target

Given P(E1) = `9/(9 + 10) = 9/19`

P(E2) = `10/(9 + 10) = 10/19`

P`("B"/"E"_1) = 12/19` ........`[∵ 1 - 7/19 = 12/19]`

P`("B"/"E"_2) = 7/19`

∴ `"P"("E"_2/"B") = ("P"("E"_2) xx "P"("B"/"E"_2))/("P"("E"_1) xx "P"("B"/"E"_1) + "P"("E"_2) xx "P"("B"/"E"_2))`

= `(10/19 xx 7/19)/(9/19 xx 12/19 + 10/19 xx 7/19)`

= `70/(108 + 70)`

= `70/178`

= 0.393

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  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 8: Descriptive Statistics and Probability - Miscellaneous Problems [पृष्ठ २०६]

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सामाचीर कलवी Business Mathematics and Statistics [English] Class 11 TN Board
पाठ 8 Descriptive Statistics and Probability
Miscellaneous Problems | Q 10 | पृष्ठ २०६
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