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جستجو

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جستجوهای پرتکرار

میتونی لایو بذاری!

A fair five-sided spinner has sides numbered 1, 2, 3, 4, 5. Raj spins the spinner and throws two fair dice. He calculates his score as follows.

- If the spinner lands on an even-numbered side, Raj multiplies the two numbers showing on the dice to get his score.

- If the spinner lands on an odd-numbered side, Raj adds the numbers showing on the dice to get his score.

Given that Raj’s score is 12, find the probability that the spinner landed on an even-numbered side.

پاسخ تشریحی :
نمایش پاسخ

$P$($E$ and $12$) $ = \frac{2}{5} \times \frac{4}{{36}} = \frac{8}{{180}}\left( {2/45} \right)$

$P\left( {12} \right) = \frac{3}{5} \times \frac{1}{{36}} + \frac{8}{{180}} = \frac{{11}}{{180}}\left( {0.0611} \right)$

$P\left( {E|12} \right) = \frac{{P\,and\,12}}{{P\left( {12} \right)}}$

$ = \frac{8}{{11}}\left( {0.727} \right)$

OR list

Even: 2 and (4,3) or (3,4) or (2,6) or (6,2)

4 and ditto

Gives 8 options

Odd: 1 and (6,6) or 3 and (6,6) or 5 and (6,6)

Gives 3 options

$\Pr ob\left( {E|12} \right) = 8/11$

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