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Hello students.
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Today we will discuss about this question.
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In this question we are given that the probability of component from a production line being defective.
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So here the probability of defective components that is equals to 0 .05 that are given and we need to find the probability that in sample of 10 components, there are two that are defective.
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That is equal to question mark.
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And in part b, we need to use the suitable approximation to find the probability that in sample of 100 components, there are 5 are defective.
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That is equals to question mark.
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So here, first of all, for the part a, we are given that number of samples that is equals to 10, that is equals to n.
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Now, probability of defective components, p that is equals to 0 .05, and the probability of non -defective components, that is 1 .5, that is equal to 1 minus 0 .05.
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That is equals to 0 .95.
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Now, the binomial distribution will be used.
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So, probability of x that is equals to r, that is equals to n, c, r p raised to r, q raise to n minus r.
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So therefore, here we can write probability of x is equals to 2, that is equals to 10c2, multiplied by 0 .05 raise to 2, 0 .95 raised to 8...