Question

Find the probability that in 10 throws of a fair die a score which is a multiple of 3 will be obtained in at least 8 of the throws.​

Answer 1

$\large\underline\blue{\bold{Given \:  Question :-  }}$

Find the probability that in 10 throws of a fair die a score which is a multiple of 3 will be obtained in at least 8 of the throws.

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$\huge{AηsωeR} ✍$

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☆Given that in a fair throw of dice, there are 10 throws.

☆We have to find the probability of multiple of 3 atleast 8 times in 10 throws.

☆We know, when a dice is rolled, the sample space is

S = {1, 2, 3, 4, 5, 6}.

☆ Let p be the probability of getting a multiple of 3 in single throw .

☆ Let q be the probability of not getting a multiple of 3.

$\sf \:  ⟼ So, \: Probability \: of \: success, p\: = \: \dfrac{2}{6} = \dfrac{1}{3}$

$\sf \:  ⟼Hence, \: the \: probability \: of \: failure \: q = 1 - p$

$\sf \:  ⟼q = 1 - \dfrac{1}{3} = \dfrac{2}{3}$

☆ Now we know that probability of 'r' successes out of 'n' trials, using binomial distribution is given by

$\bf \:  ⟼ P(r) = \: ^{n} C_r {p}^{r} {q}^{n - r}$

where,

• ¤ n is number of trials
• ¤ p is probability of success
• ¤ q is probability of failure

$\begin{gathered}\begin{gathered}\bf here : \begin{cases} &\sf{n \: = 10} \\ &\sf{p = \dfrac{1}{3} }\\ &\sf{q = \dfrac{2}{3} } \end{cases}\end{gathered}\end{gathered}$

☆ Now, we have to calculate the probability of multiple of 3 getting atleast 8 times in 10 throws, is given by

$\sf \:  ⟼ \: P(8) + P(9) + P(10)$

$\sf \:  = \:^{10} C_8 {\bigg(\dfrac{1}{3} \bigg)}^{8} {\bigg( \dfrac{2}{3} \bigg)}^{2} + \:^{10} C_9 {\bigg(\dfrac{1}{3} \bigg)}^{9} {\bigg( \dfrac{2}{3} \bigg)}^{1} + \:^{10} C_10 {\bigg(\dfrac{1}{3} \bigg)}^{10} {\bigg( \dfrac{2}{3} \bigg)}^{0}$

$\sf \:  = \dfrac{10 \times 9}{2 \times 1} \times \dfrac{4}{ {3}^{10} } + 10 \times \dfrac{2}{ {3}^{10} } + \dfrac{1}{ {3}^{10} }$

$\sf \:  = \dfrac{1}{ {3}^{10} } (180 + 20 + 1)$

$\sf \:  = \: \dfrac{201}{ {3}^{10} }$

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