Probability

The word “MUMBAI” has six letters. Therefore, total ways of arrangements = 6!/2 =360. Number of words where two M’s do not come together is
= Total words – Number of words where they come together
= 360 – 5! = 240
Hence, the required probability is = 240/360 = 2/3.

Total number of exhaustive cases = 6 × 6 × 6 = 216. Cases favourable to get a total of 7 is = 12 that is, (1,1,5); (1,2,4); (1,3,3); (1,4,2); (1,5,1); (2,1,4); (2,2,3); (2,3,2); (2,4,1); (3,1,3); (3,2,2); (3,3,1); ( 4,1,2); (4,2,1); (5,1,1). Hence, the required probability is = 15 /216 = 5/72.

There are seven perfect square numbers in the given range. Let A = Event of getting a number, which is a square = ( 1,4,9,16,25,36, 49). Hence, n(A) = 7. There are total 50 cards, so 50 possible outcomes. Hence, n(S) = 50. Therefore, required probability, P(A) = n(A) / n(S) = 7/50.

Probability of getting a white ball from first urn is, P(W₁)=4/6 = 2/3. Probability of getting a white ball from second urn is, P(W₂)= 5/8. Therefore, the required probability that both the balls are white is = 2/3 × 5/8 = 5/12.

Probability of getting a digital watch out of the first bag is = 1/2 x 3/8 = 3/16. Probability of getting a digital watch out of the second bag is = 1/2 x 4/11 = 2/11. Both the two events are independent event. Therefore, required probability is, P = (1/2 x 3/8 + 1/2 x 4/11) = 65/ 176.

A pack of playing card contains 52 cards, divided into 4 suits. These are – Spades, Clubs, Hearts and Diamonds. Each suit has 13 cards. Among these 13 cards, 9 cards are numbered 2 to 10. The remaining four cards are ace, king, queen and jack. Therefore, each suit contains one ace. Hence, there are total 4 aces in the pack. Out of 4 aces, two can be drawn in ⁴C₂ ways. Here, n(E) = ⁴C₂ = 6. Again, from a pack of 52 cards, 2 cards can be drawn in ⁵²C₂ ways.Therefore, n(S) = ⁵²C₂ = 1326 and the required probability is 6/1326 = 1/221.

Number of ways in which six letters can be arranged is 6!/2 = 720/2 = 360 ways. If N always comes next to A, they both can be treated as a single letter. In that case, the total arrangements will be, 5!. Therefore, the required probability is 5! / 360 = 1/3.

A pack of playing card contains 52 cards, divided into 4 suits. These are – Spades, Clubs, Hearts and Diamonds. Each suit has 13 cards. Among these 13 cards, 9 cards are numbered 2 to 10. The remaining four cards are ace, king, queen and jack. From a pack of 52 cards, 3 cards can be drawn in ⁵²C₃ ways. Therefore, the required probability is, (⁴C₁×⁴C₁׳⁶C₁)/⁵²C₃ = 144/5525.

Let, A = probability of first event taking place, P(A) = 3/(7 + 3) = 3/10. B = probability of second event taking place, P(B) = 5/(5 + 2) = 5/7. The event will occur if  either only A takes place and B does not take place or B takes place and A does not take place, or both the events take place. Therefore, the required probability will be = [P(A) x {1 - P(B)}] + [P(B) x {1 - P(A)}] + [P(A) x P(B)] = [3/10 x (1 - 5/7)] + [5/7 x (1 - 3/10)] + [3/10 x 5/7] = 4/5.

Probability that John can solve the problem is 1/3. Probability that John cannot solve the problem is (1 - 1/3) = 2/3. Probability that Johny can solve the problem is 1/5. Probability that Johny cannot solve the problem is (1-1/5) = 4/5. Probability that Janet can solve the problem is 1/6. Probability that Janet cannot solve the problem is (1 - 1/6) = 5/6. The probability that they together cannot solve the problem is = 2/3 × 4/5 × 5/6 = 4/9. Therefore, the required probability that the problem will be solved is = (1 - 4/9) = 5/9.

Let, P(R) = Probability of John’s selection = 1/5.  So, the probability that John is not selected is P(R ̅) = 1 - 1/5 = 4/5. Similarly, P(A) = Probability of Janet’s selection = 1/7. So, P(A ̅) = 1 - 1/7 = 6/7. Hence, the probability that only one of them will be selected is = P(R) × P(A ̅) + P(A) × P(R ̅) = 2/7.

The product will be even if both or at least one ticket is even numbered. There are 25 even and 25 odd numbered tickets. Number of ways of selecting even numbered tickets = ²⁵C_2. Number of ways of selecting one even and one odd numbered ticket is = ²⁵C₁ × ²⁵C_1. From 50 tickets, 2 tickets can be drawn in ⁵⁰C₂ ways.Therefore, the required probability is = (²⁵C₂+²⁵C₁ × ²⁵C₁)/⁵⁰C₂ = 37/49.

Let R: John speaks the truth; and A: Janet speaks the truth. Hence, P(R) = 70/100 and P (R ̅) = 1 - 70/100 = 30/100. Again, P (A) = 80 /100 and P (A ̅) = 1 - 80/100 = 20/100. Therefore, required probability,P(R) × P (A ̅) + P (A) × P (R ̅) = 70/100 x 20/100 + 80/100 x 30/100 = 19/ 50

There are 5 letters. The probability, that the letter I gets the first position is 1/5. The probability that M is in the second position is 1/4. Similarly, the probability for A, G and E are 1/3 , 1/2, 1/1,  respectively. Hence, the required probability is = (1/5 × 1/4 × 1/3 × 1/2 × 1/1 ) = 1/120.

A pack of playing card contains 52 cards, divided into 4 suits. These are – Spades, Clubs, Hearts and Diamonds. Each suit has 13 cards. Among these 13 cards, 9 cards are numbered 2 to 10. The remaining four cards are ace, king, queen and jack. From a pack of 52 cards, 2 cards can be drawn in ⁵²C_² ways. There may be two different events - Let, A = event of selecting both cards from club suit. Here, one card is king of club and B= event of selecting a card from club suit and a non-club king. Therefore, the probability will be –[(¹²C₁ x ¹C₁)/⁵²C₂] + [(¹³C₁ x ³C₁)/⁵²C₂] = 1/26.