A die is thrown 6 times. If ‘getting an odd number’ is a success,

1.	A die is thrown 6 times. If ‘getting an odd number’ is a success, what is the probability of (i) 5 successes? (ii) at least 5 successes? (iii) at most 5 successes?
Answer» The repeated tosses of a die are Bernoulli trials. Let X denote the number of successes of getting odd numbers in an experiment of 6 trials. p = P (success) = P (getting an odd number in a single throw of a die) ∴ p= 3/6 = 1/2 and q = P (failure) = 1 - p = 1 - 1/2 = 1/2 Therefore, by Binomial distribution P(x = r) = ⁿC_rp^n-rq^r, where r = 0, 1, 2,...., n P(X = r) = ⁶C_r(1/2)^6-r(1/2)^r = ⁶C_r(1/2)⁶ (i) P(5 successes)= ⁶C₅(1/2)⁶ = 6 x 1/2⁶ = 6/64 = 3/32 (ii) P (at least 5 successes) = P (5 successes) + P (6 successes) =⁶C₅(1/2)⁶ + ⁶C₆(1/2)⁶ = 6/64 + 1/64 = 7/64 (iii) P (at most 5 successes) = 1 – P (6 successes) = 1 - ⁶C₆(1/2)⁶ = 1 - 1/64 = 63/64

A die is thrown 6 times. If ‘getting an odd number’ is a success, what is the probability of (i) 5 successes? (ii) at least 5 successes? (iii) at most 5 successes?

Answer»

The repeated tosses of a die are Bernoulli trials.

Let X denote the number of successes of getting odd numbers in an experiment of 6 trials.

p = P (success) = P (getting an odd number in a single throw of a die)

∴ p= 3/6 = 1/2 and q = P (failure) = 1 - p = 1 - 1/2 = 1/2

Therefore, by Binomial distribution

P(x = r) = ⁿC_rp^n-rq^r, where r = 0, 1, 2,...., n

P(X = r) = ⁶C_r(1/2)^6-r(1/2)^r = ⁶C_r(1/2)⁶

(i) P(5 successes)= ⁶C₅(1/2)⁶ = 6 x 1/2⁶ = 6/64 = 3/32

(ii) P (at least 5 successes) = P (5 successes) + P (6 successes) =⁶C₅(1/2)⁶ + ⁶C₆(1/2)⁶ = 6/64 + 1/64 = 7/64

(iii) P (at most 5 successes) = 1 – P (6 successes) = 1 - ⁶C₆(1/2)⁶ = 1 - 1/64 = 63/64

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