Multiple choice

A die is thrown $7$ times. What is the chance that an odd number turns up ? (i) exactly $4$ times (ii) at least $4$ times.

  1. (i)$\displaystyle \frac{35}{128}$, (ii)$\displaystyle \frac{1}{2}$
  2. (i)$\displaystyle \frac{45}{128}$, (ii)$\displaystyle \frac{1}{2}$
  3. (i)$\displaystyle \frac{35}{128}$, (ii)$\displaystyle \frac{1}{4}$

  4. None of these
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A Correct answer
Explanation

Probability of odd number (p) = 1/2. (i) P(X=4) = 7C4 * (1/2)^4 * (1/2)^3 = 35 * (1/128) = 35/128. (ii) P(X>=4) = P(4)+P(5)+P(6)+P(7) = (35+21+7+1)/128 = 64/128 = 1/2.

AI explanation

The probability of getting an odd number on a die is p = 3/6 = 1/2, and the probability of an even number is q = 1/2. Using the binomial probability formula for exactly 4 odd numbers in 7 throws, the probability is 7C4 * (1/2)^4 * (1/2)^3 = 35 * (1/16) * (1/8) = 35/128. For at least 4 odd numbers, we sum the probabilities for 4, 5, 6, and 7 successes using the binomial expansion of (1/2 + 1/2)^7, yielding (35 + 21 + 7 + 1) / 128 = 64/128 = 1/2. The required chances are 35/128 and 1/2.

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