An unbalanced coin is weighted so that the probability of heads is 0.55. The coin is tossed ten times. (a) What is the probability of getting exactly 6 heads? (b) What is the probability of getting fewer than 3 heads?

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An unbalanced coin is weighted so that the probability of heads is 0.55. The coin is tossed ten times.   (a) What is the probability of getting exactly 6 heads?   (b) What is the probability of getting fewer than 3 heads?

Fearen · Accepted Answer

Step 1 Probability quantifies the chances of happening an event. The probability values always lie in the range 0 and 1. Probability distributions are of two types, discrete and continuous distributions. Binomial distribution is discrete distribution that deals with the probabilities of success and failure. Step 2 Let the random variable X indicates the number of heads obtained on tossing a coin 10 times with probability of head is 0.55 (unbiased coin). So the number of trials is fixed and each trial has 2 possible outcomes and each trial is independent. The probability of success is fixed across all the trials. Thus the X is binomial random variable. X∼Binom(10,0.55) a) P(X=6) indicates the probability of getting exactly 6 heads. This can be calculated in the following way. P(X=x)=(nx)px(1−p)n−x P(X=6)=(106)0.556×(1−0.55)4 =10!6!(10−6)!0.556(0.45)4 =0.2384 Thus 0.2384 is the required probability. Step 3 b) P(X&amp;lt;3) denotes teh probability of getting less than 3 heads and this can be calculated in the following way. P(X&amp;lt;3)=P(X=0)+P(X=1)+P(X=2) =(100)0.550×(1−0.55)0+(101)0.551×(1−0.55)1+(102)0.552×(1−0.55)2 =10!0!(10−0)!0.550(0.45)10+10!1!(10−1)!0.551(0.45)9+10!2!(10−2)!0.552(0.45)8 =0.0003+0.004+0.023 =0.0273 Thus 0.0273 is the required probability.

An unbalanced coin is weighted so that the probability of

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