Question

2) Following table shows there are 20 nails in one box.
Number of nails | 30 | 32 | 28 | 33 | 31
.Number of boxes
2 1 4
5 1 6 3
If one box is selected randomly, Find the probability of following events.
i) 31 nails in box ii) Less than 33 nails in box
M. at least 28 nails in box iv) more than 33 nails in box.​

Answer 1

1) 31 nails in box

Solution :

S = {2+4+5+6+3}

So n(S) = 20

Event is 31 nails. That means boxes which have nails greater than 31 to be considered.

A = { 4+6+3}

So n(A) = 13

Therefore,

P(A) = n(A) /n(S)

= 13/20

2) Less than 33 nails in box

n(S) = 20

A = { 2+5+4+3}

So n(A) = 14

Therfore,

P(A) = n(A) /n(S)

= 14/20

= 7/10

3) At least 28 nails in box

n(S) = 20

A = { 5+2+4+6+3}

So n(A) = 20

Therefore,

P(A) = n(A) / n(S)

= 20/20

= 1

4) More than 33 nails in the box

n(S) = 20

A = { 0}

So n(A) = 0

Therefore,

P(A) = n(A) /n(S)

= 0/20

= 0