Question

A box contains 100 bolts and 50 nuts. It is given that 40% bolts and 60%nuts are rusted. Two object are selected from the box at random. Find the probability that either both are bolt or bolt are rusted

Answer 1

Answer:

66/149

491/745

Step-by-step explanation:

Question says Find the probability that either both are bolt or bolt are rusted

bolt are rusted - should be both are rusted ( we will solve both cases)

Case 1 -  Both are bolts or bolts rusted

100 Bolts + 50 nuts

Probability of 1st bolt =  100/(100+50) = 100/150 = 2/3

1 bolt has been taken so now 100-1 = 99 bolts available & 50 nuts available

Probability of 2nd Bolt = 99/(99+50) = 99/149

Probability of being both bolts = (2/3) * (99/149) = 66/149

if we assume question is correct  and mentioned additionally that or bolt is rusted

now Rusted bolt is subset of bolts and mentioned or in question so probability remains same = 66/149

Case 2 - both are bolt or both are rusted

Probability of being both bolts = 66/149 already calculated

40% bolts are rusted = (40/100) * 100 = 40 Bolts rusted

60% nuts are rusted = (60/100) * 50 = 30 Nuts Rusted

so total 40 + 30 = 70 rusted

Probability of 1st Rusted =  70/(100+50) = 70/150 = 7/15

1 Rusted  has been taken so now 70-1 = 69 Rusted available and total 150-1 = 149 available

Probability of 2nd Rusted = 969/(149) = 69/149

Probability of being both rusted = (7/15) * (69/149)

= (7/5) * (23/149)

= 161/745

Probability of both are bolt or both are rusted

= 66/149  + 161/745

= (330 + 161)/745

= 491/745

Answer 2

Answer:

Step-by-step explanation:

Number of bolts in the box = 100

Number of nuts in the box = 50

Number of rusted bolts = 50

Number of rusted nuts = 25

Total number of items in the box = 100+50 = 150

Total number of rusted items in the box = 50+ 25 = 75

Consider the following events : A = Event of getting a rusted item, B = Event of getting a bolt and A ∩ B = Event of getting a rusted bolt.

Number of ways of drawing 2 items out of 150 = 150c₂

There are 75 rusted items, out of which 2 can be drawn in 75c₂ ways

So, P (A) = 75c₂/150c₂ = (75*74/2*1)/(150*149/2*1)

P (A) = 2775/11175

There are 100 bolts in the box, out of which 2 can be drawn in 100c₂ ways.

So, P (B) = 100c₂/150c₂ = (100*99/2*1)/(150*149/2*1)

P (B) = 4950/11175

There are 50 bolts that are rusted, out of which 2 can be drawn in 50c₂ ways.

P (A ∩ B) = 50c₂/150c₂ = (50*49/2*1)/(150*149/2*1)

P (A ∩ B) = 1225/11175

So, the required probability = P (A ∪ B) = P (A) + P (B) - P (A ∩ B)

= 2775/11175 + 4950/11175 - 1225/11175

= (7725 - 1225)/11175

= 6500/11175

P (A ∪ B) = 260/447

Hope it's help