Algebra

Probability

Medium

applied

A die is thrown 6 times. If getting an odd number is a success, then the probability of at least 5 successes is

$\frac{63}{64}$

$\frac{9}{64}$

$\frac{7}{64}$

$\frac{1}{64}$

✅ Correct Option: 3

A die has 6 faces: 1, 2, 3, 4, 5, 6

Odd numbers: 1, 3, 5 (3 odd numbers)

Even numbers: 2, 4, 6 (3 even numbers)

Probability of success (getting an odd number): $p = \frac{3}{6} = \frac{1}{2}$

Probability of failure (getting an even number): $q = \frac{3}{6} = \frac{1}{2}$

"At least 5 successes" means exactly 5 successes or exactly 6 successes.

$P(\text{at least 5}) = P(\text{exactly 5}) + P(\text{exactly 6})$

The binomial probability formula is:

$P(X = r) = \binom{n}{r} \times p^r \times q^{n-r}$

where $n = 6$ , $p = \frac{1}{2}$ , $q = \frac{1}{2}$

For exactly 5 successes:

$P(X = 5) = \binom{6}{5} \times \left(\frac{1}{2}\right)^5 \times \left(\frac{1}{2}\right)^{1}$

$= 6 \times \frac{1}{32} \times \frac{1}{2}$

$= 6 \times \frac{1}{64}$

$= \frac{6}{64}$

For exactly 6 successes:

$P(X = 6) = \binom{6}{6} \times \left(\frac{1}{2}\right)^6 \times \left(\frac{1}{2}\right)^{0}$

$= 1 \times \frac{1}{64} \times 1$

$= \frac{1}{64}$

$P(\text{at least 5 successes}) = P(X = 5) + P(X = 6)$

$= \frac{6}{64} + \frac{1}{64}$

$= \frac{7}{64}$

Therefore, the probability of at least 5 successes is $\frac{7}{64}$ .

2025: 21 May Shift 1

There are 50 telephone lines in an exchange. The probability that any one of them will be busy is 0.1, then the probability that all the lines are busy?

2025: 22 May Shift 2

Bag I contains 3 black and 2 white balls. Bag II contains 2 black and 4 white balls. A bag is selected at random and then a ball is drawn from it. The probability that the ball drawn is black is:

2025: 3 June Shift 2

The probabilities of occurrence of two events E and F are 0.25 and 0.50 respectively. The probability of their simultaneous occurrence is 0.14. The probability that neither E nor F occurs is

2025: 15 May Shift 2

A player participates in 3 matches against three teams T₁, T₂ and T₃.The probability of winning a match against teams T₁, T₂ and T₃ are 0.2, 0.3 and 0.9 respectively. If 'wins' can be regarded as independent events, then the probability that he

(A) wins all the 3 matches is 0.054

(B) wins no match is 0.054

(D) wins exactly one match is 0.542

Choose the correct answer from the options given below:

2022: 17 Aug Shift 2

The random variable $X$ has a probability distribution

$P(X=x) = \begin{cases} 5k, & x=0 \\ 2k, & x=1 \\ 3k, & x=2 \\ 0, & \text{otherwise.} \end{cases}$

Then, the value of $E(X)$ is :

2022: 25 May Shift 1

Two dice are thrown simultaneously. If X denotes the number of sixes, then the variance of X is:

A die is thrown 6 times. If getting an odd number is a success, then the probability of at least 5 successes is

Solution

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