Question

80 bulbs are selected at random from a lot and their life time (in hours) is

recorded in the form of a frequency table given below

One bulb is select from lot

a. Probability of that its life is 1150 hours.

b. Probability of that life is more that 500 hours.

c. Probability of that life is atleast 500 hours​

Answer 1

Required Answer:-

The required frequency distribution table missing in ur question is in the above attachment. Regarding that, Let's solve these questions.

We know that:

Probability of any event is defined as the no. of favourable outcomes / no. of total outcomes.

$\huge{ \boxed{ \bf{(a)}}}$

• No. of bulbs with life 1150 h = 0

• Total no. of bulbs

= 10 + 12 + 23 + 25 + 10

= 80

• Then, P(Bulb with life 1150 h)

= 0/80

= 0

$\huge{ \boxed{ \bf{(b)}}}$

• No. of bulbs with life > 500 h

= 23 + 25 + 10

= 58

• Total no. of bulbs = 80

• Then, P(more than 500 h)

= 58/90

= 29/45

$\huge{ \boxed{ \bf{(c)}}}$

• No. of bulbs with life atleast 500 h

= 12 + 23 + 25 + 10

= 70

• Total no. of bulbs = 80

• Then, P(atleast 500 h)

= 70/80

= 7/8

Note:

• When it is given atleast (a number) means equal to that number and more than that.

• When it is given atmost then, the number equal or less than that number.

Answer 2

Required Answer :-

Answer (1)

In the frequency table there is no bulb with the life 1150 hours.

Therefore,

As we know that

$\star {\boxed {\green {\underline { \bf \: P = \frac{favourable \: no. \: of \: outcomes}{total \: outcomes} }}}}$

10 + 12 + 23 + 25 + 10 = 80

Therefore

P = 0/80

Answer (2)

23 + 25 + 10

= 58

• Total no. of bulbs = 80.

58/90

29/45

Answer (3)

Total bulb = 12 + 23 + 25 + 10 = 70

Then, P(atleast 500 h)

70/80

7/8