Probability - Study Mode

[#56] A box contains 100 pens. Out of which eight are defective. One pen is out from the box. Find the probability that the pen is not defective.
Correct Answer

(A) $$frac{{23}}{{25}}$$

Explanation

Solution: Good pen = 100 - 8 = 92 Probability = $$frac{{92}}{{100}}$$ = $$frac{{23}}{{25}}$$

[#57] When two coins are tossed simultaneously, what are the chances of getting at least one tail?
Correct Answer

(C) $$frac{{3}}{{4}}$$

Explanation

Solution: At least one tails means one tails or both tails Let us see what the possible outcomes are when two coins are tossed simultaneously We can get, 1. Heads and Heads 2. Heads and Tails 3. Tails and Heads 4. Tails and Tails So we have total = 4 outcomes Out of them, 3 have tails in it Chances of at least 1 tails = $$frac{{3}}{{4}}$$

[#58] A box has 5 black and 3 green shirts. One shirt is picked randomly and put in another box. The second box has 3 black and 5 green shirts. Now a shirt is picked from second box. What is the probability of it being a black shirt?
Correct Answer

(B) $$frac{{29}}{{72}}$$

Explanation

Solution: From box 1 we can pick black or green shirt Case 1: Pick black shirt Box 1 has total 5 + 3 = 8 shirts Probability of black from box 1 = $$frac{{5}}{{8}}$$ Now this black is added to box 2 So box 2 now has 3 + 1 = 4 black & 5 green shirts Total = 4 + 5 = 9 shirts Probability of black from box 2 = $$frac{{4}}{{9}}$$ Case 1 probability $$eqalign{
& = frac{5}{8} imes frac{4}{9} cr
& = frac{{20}}{{72}} cr} $$ Case 2: Pick green shirt Probability of green from box 1 = $$frac{{3}}{{8}}$$ Now this green is added to box 2 So box 2 now has 3 black & 5 + 1 = 6 green shirts Total = 3 + 6 = 9 shirts Probability of black from box 2 = $$frac{{3}}{{9}}$$ Case 2 probability $$eqalign{
& = frac{3}{8} imes frac{3}{9} cr
& = frac{9}{{72}} cr} $$ Total Probability $$eqalign{
& = frac{{20}}{{72}} + frac{9}{{72}} cr
& = frac{{29}}{{72}} cr} $$

[#59] In a set of 30 game cards, 17 are white and rest are green. 4 white and 5 green are marked IMPORTANT. If a card is chosen randomly from this set, what is the possibility of choosing a green card or an ‘IMPORTANT’ card?
Correct Answer

(C) $$frac{{17}}{{30}}$$

Explanation

Solution: We want green card or IMPORTANT card There are 30 - 17 = 13 green cards There are 4 + 5 = 9 IMPORTANT cards Total cards = 30 Also 5 green cards are IMPORTANT cards So probability $$eqalign{
& = frac{{13}}{{30}} + frac{9}{{30}} - frac{5}{{30}} cr
& = frac{{17}}{{30}} cr} $$

[#60] Suresh keeps all his socks in a single drawer. He has 24 pairs of white socks and 18 pairs of grey socks. Suresh picks 3 socks randomly. Find the possibility of Suresh choosing a matching pair.
Correct Answer

(D) 1

Explanation

Solution: Since a pair has 2 socks in it and there are just two colors - white and grey, when Suresh chooses 3 socks, he will surely select 2 socks of same color - be it grey or white. So possibility = probability = 1