If a box contains 10 bulbs,of which just three are defective. If a random sample of five bulbs is drawn, find the probability that the sample contains no defective bulbs.

Correct Answer: 2/7

Explanation:

P( only one of them will be selected) = p[(E and not F) or (F and not E)] 

 =  P E ? F ? F ? E 

 

=  P E P F + P F P E

 

 = 1 7 × 4 5 + 1 5 × 6 7 = 2 7

Correct Answer: 91/256

Explanation:

Find the number of cases in which none of the digits show a '6'.

i.e. all three dice show a number other than '6', 5×5×5=125 cases.

Total possible outcomes when three dice are thrown = 216.

The number of outcomes in which at least one die shows a '6' = Total possible outcomes when three dice are thrown - Number of outcomes in which none of them show '6'.

=216?125=91

The required probability = 91/256

Correct Answer: 0

Explanation:

The probability of an impossible event is 0.

The event is known ahead of time to be not possible, therefore by definition in mathematics, the probability is defined to be 0 which means it can never happen.

 

The probability of a certain event is 1.

Correct Answer: 1/5

Explanation:

Correct Answer: 1/2

Explanation:

The possible outcomes are as follows :

5H, 5T, (H, 4T), (T, 4H), (2H, 3T) (3H, 2T), i.e. 6 outcomes in all.

Therefore the probability that head appears an odd number of times = 3/6 =1/2 (In only three outcomes out of the six outcomes, head appears an odd number of times).

Correct Answer: 25/34

Explanation:

The events of selection of two person is redefined as first is a girl and second is a boy or first is boy and second is a girl or first is a girl and second is a girl.

So the required probability:
= 8 17 * 9 16 + 9 17 * 8 16 + 8 17 * 7 16

 

=  9 34 + 9 34 + 7 34
=  25 34

Correct Answer: 1/7

Explanation:

Given total number of smileys = 5 + 6 + 3 = 14

Now, required probability = 

$$\begin{gathered} 5 C 2 14 C 2 + 3 C 2 14 C 2 \\ = 10 91 + 3 91 \\ = 13 91 \\ = 1 7 \end{gathered}$$

Correct Answer: 1

Correct Answer: 1/3

Explanation:

Here Sample space S = { HH, HT, T1, T2, T3, T4, T5, T6 }

 

Let A be the event that the die shows a number greater than 4 and B be the event that the first throw of the coin results in a tail then,

 

 A = { T5, T6 }

 

 B = { T1, T2, T3, T4, T5, T6 }

 

Therefore, Required probability =  P A B = P A ? B P B = 2 8 6 8 = 1 3

Correct Answer: 14/33

Explanation:

Total number of chairs = (3 + 5 + 4) = 12.

Let S be the sample space.

Then, n(s)= Number of ways of picking 2 chairs out of 12

= 12×11/2×1 = 66

Let n(E) = number of events of selecting 2 chairs for selecting no white chairs.

=> 8C2 = 8×7/2×1 = 28

Therefore required probability = 28/66 = 14/33.