Math, asked by anandkm4064, 11 months ago

A die is thrown 10 times. If getting a prime number is considered a success, find the probability of getting not more than 8 successes.

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Answered by gj12345567
1

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Answered by amirgraveiens
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The probability of getting not more than 8 successes is 0.896.

Step-by-step explanation:

Given:

Number of times die is thrown, n=10

The numbers on a die are 1, 2, 3, 4, 5 and 6.

The numbers that are prime are 1, 2, 3, and 5.

Therefore, probability of getting a prime number is given as:

P(Prime)=\frac{\textrm{Number of prime numbers}}{Total\ outcomes}\\\\P(Prime)=\frac{4}{6}=\frac{2\times 2}{2\times 3}=\frac{2}{3}

Probability of getting a success is, p=\frac{2}{3}

So, probability of getting a failure is, q=1-p=1-\frac{2}{3}=\frac{3-2}{3}=\frac{1}{3}

As per Bernoulli's theorem of getting 'x' successes out of 'n' trials is given as:

P(X=x)=_{x}^{n}\textrm{C}p^xq^{n-x}

Probability of getting not more than 8 successes can be determined using Bernoulli's theorem as:

Number of trials is, n=10

Number of successes is, X\leq 8

Therefore, P(X\leq 8)=1-[P(X=9)+P(X=10)]\\P(X\leq 8)=1-[_{9}^{10}\textrm{C}(\frac{2}{3})^{9}(\frac{1}{3})^{1}+_{10}^{10}\textrm{C}(\frac{2}{3})^{10}]\\P(X\leq 8)=1-[10\times 0.026\times 0.333+1\times 0.017]\\P(X\leq 8)=1-[0.087+0.017]\\P(X\leq 8)=1-0.104=0.896

Therefore, the probability of getting not more than 8 successes is 0.896.

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