A pair of dice is thrown 5 times. If getting a total of 7 is considered a success, then
Match the LIST-I with LIST-II
| LIST-I | LIST-II | ||
| A. | Probability of no success | I. | \( 26\left(\displaystyle\frac{1}{6}\right)^{5} \) |
| B. | Probability of 5 successes | II. | \( 2\left(\displaystyle\frac{5}{6}\right)^{5} \) |
| C. | Probability of at least 4 successes | III. | \( \left(\displaystyle\frac{1}{6}\right)^{5} \) |
| D. | Probability of at most one success | IV. | \( \left(\displaystyle\frac{5}{6}\right)^{5} \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → A-IV, B-III, C-I, D-II
A total of 7 occurs in 6 of the 36 outcomes, so on one throw
\( \displaystyle p=\frac{6}{36} \)
\( \displaystyle =\frac{1}{6} \)
\( \displaystyle q=\frac{5}{6} \)
The number of successes in 5 throws is binomial, with \( P(X=r)={}^{5}C_{r}\,p^{r}q^{5-r} \).
(A) No success:
\( \displaystyle P(X=0) \)
\( \displaystyle =\left(\frac{5}{6}\right)^{5} \)
This matches (IV).
(B) 5 successes:
\( \displaystyle P(X=5) \)
\( \displaystyle =\left(\frac{1}{6}\right)^{5} \)
This matches (III).
(C) At least 4 successes:
\( \displaystyle P(X=4)+P(X=5) \)
\( \displaystyle =5\left(\frac{1}{6}\right)^{4}\left(\frac{5}{6}\right)+\left(\frac{1}{6}\right)^{5} \)
\( \displaystyle =\frac{25+1}{6^{5}} \)
\( \displaystyle =26\left(\frac{1}{6}\right)^{5} \)
This matches (I).
(D) At most one success:
\( \displaystyle P(X=0)+P(X=1) \)
\( \displaystyle =\left(\frac{5}{6}\right)^{5}+5\cdot\frac{1}{6}\left(\frac{5}{6}\right)^{4} \)
\( \displaystyle =\left(\frac{5}{6}\right)^{4}\left(\frac{5}{6}+\frac{5}{6}\right) \)
\( \displaystyle =2\left(\frac{5}{6}\right)^{5} \)
This matches (II).
So the matching is (A)-(IV), (B)-(III), (C)-(I), (D)-(II).