Question

Suppose that 90% of people are right-handed. What is the probability that at most 6 of a random sample of 10 people are right-handed?

Answer 1

3/5
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Answer 2

Answer:

1 - (9⁷/10¹⁰) ( 2064)

0.0128

Step-by-step explanation:

p = 9/10  probability of right handed

q = 1 - 9/10 = 1/10  probability of left handed ( or not right handed)

probability that at most 6 of a random sample of 10 people are right-handed

= 1 - Probability of 7 , 8 , 9 & 10 people are right handed

= 1  -  ¹⁰C₇p⁷q³ - ¹⁰C₈p⁸q² - ¹⁰C₉p⁹q¹ - ¹⁰C₁₀p¹⁰q⁰

= 1 - ( 120 * 9⁷/10¹⁰ + 45 * 9⁸/10¹⁰ + 10 * 9⁹/10¹⁰ + 1 * 9¹⁰/10¹⁰)

= 1 - (9⁷/10¹⁰) ( 120 + 45*9 + 10*9² + 9³)

= 1 - (9⁷/10¹⁰) ( 2064)

= 1 - 0.9872

= 0.0128